Chapter 7
Chapter 7 Maintaining Mathematical Prociency (p. 335)
12. 30 = 2
77 = 7
1. 3x  7 + 2x = 3x + 2x  7
⋅3⋅5
⋅ 11
The numbers have no common prime factors. So, the GCF of
30 and 77 is 1.
= (3 + 2)x  7
= 5x  7
2. 4r + 6  9r  1 = 4r  9r + 6  1
= (4  9)r + 6  1
13. It is not possible for two integers to have no common factors
because 1 is a factor of every integer.
= 5r + 5
Chapter 7 Mathematical Practices (p. 356)
3. 5t + 3  t  4 + 8t = 5t  t + 8t + 3  4
1. The algebraic expression is 3x 2  x + 1.
= (5  1 + 8)t + 3  4
= (2)t + (1)
= 2t  1
= 3s  3 + 5
5. The algebraic expression is x 2 + 2.
= 3s + 2
5. 2m  7(3  m) = 2m  7(3)  7(m)
= 2m  21 + 7m
= 9m  21
= 4h + 24  h + 2
= 4h  h + 24 + 2
= (4  1)h + 24 + 2
= 3h + 26
7. 20 = 2
⋅3⋅7
63 = 3 ⋅ 3 ⋅ 7
⋅ 2 = 4.
8. 42 = 2
The GCF of 42 and 63 is 3
⋅3⋅3⋅3
81 = 3 ⋅ 3 ⋅ 3 ⋅ 3
⋅
7 = 21.
9. 54 = 2
The GCF of 54 and 81 is 3
⋅2⋅2⋅3⋅3
84 = 2 ⋅ 2 ⋅ 3 ⋅ 7
⋅ ⋅
3
3 = 27.
2
9. The algebraic expression is 2x 2.
7.1 Explorations (p. 357)
1. Step 1
(3x + 2) + (x  5)
Step 2
4x + 2  5
Step 3
4x + (2  2)  3
Step 4
4x  3
Step 2
( x 2 + 2x + 2 )  (x  1)
( x 2 + 2x + 2 ) + (x + 1)
Step 3
x 2 + 2x  x + 3
Step 4
x 2 + x + (x  x) + 3
Step 5
x2 + x + 3
2. Step 1
the opposite of each of its terms. Use the Commutative and
Associative Properties to rearrange the terms so that like
terms are grouped together. Then add or subtract like terms.
4. a. ( x 2 + 2x  1 ) + ( 2x 2  2x + 1 )
⋅ 2 ⋅ 3 = 12.
⋅ ⋅
64 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2
The GCF of 28 and 64 is 2 ⋅ 2 = 4
11. 28 = 2
7. The algebraic expression is x 2 + 2x.
3. Sample answer: If a polynomial is being subtracted, nd
10. 72 = 2
The GCF of 72 and 84 is 2
6. The algebraic expression is x  6.
8. The expression is 9.
6. 4(h + 6)  (h  2) = 4(h) + 4(6)  h  (2)
The GCF of 20 and 36 is 2
3. The algebraic expression is x 2  2x.
4. The algebraic expression is x 2 + x  1.
4. 3(s  1) + 5 = 3(s)  3(1) + 5
⋅2⋅5
36 = 2 ⋅ 2 ⋅ 3 ⋅ 3
2. The algebraic expression is x 2 + x.
7
= x 2 + 2x 2 + 2x  2x  1 + 1
= ( x 2 + 2x2 ) + (2x  2x) + (1 + 1)
= 3x 2 + 0 + 0
= 3x 2
b. (4x + 3) + (x  2) = 4x + x + 3  2
= (4x + x) + (3  2)
= 5x + 1
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Worked-Out Solutions
379
Chapter 7
c. ( x 2 + 2 )  ( 3x 2 + 2x + 5 )
9.
= ( x 2 + 2 ) + ( 3x 2  2x  5 )
=
x2

3x 2
+ 7x 2  x
 2x + 2  5
8x 2  2x  2
= ( 1x 2  3x 2 )  2x + (2  5)
=
2x 2
x2  x  2
The sum is 8x 2  2x  2.
 2x  3
d. (2x  3x)  ( x 2  2x + 4 ) = (x) + ( x 2 + 2x  4 )
10. ( p 2 + p + 3 )  ( 4p 2  p + 3 )
= p 2 + p + 3 + 4p 2 + p  3
= x  x 2 + 2x  4
= p 2 + 4p 2 + p + p + 3  3
= x 2  x + 2x  4
= ( p 2 + 4p 2 ) + ( p + p) + (3  3)
= x 2 + (x + 2x)  4
= 5p 2 + 2p + 0
= x 2 + x  4
= 5p 2 + 2p
The difference is 5p 2 + 2p.
7.1 Monitoring Progress (pp. 358–361)
1. In the monomial 3x 4, the exponent of x is 4. So, the degree
of the monomial is 4.
k + 5
11.
( 3k 2
5
3. In the monomial —3 y, the exponent of y is 1. So, the degree of
the monomial is 1.
+6
The difference is 3k 2  k + 11.
12. a. Penny:
16t2  25t + 200
Paintbrush:  ( 16t2
16t2  25t + 200
+ 100 ) ⇒ +16t2
 100
25t + 100
4. You can rewrite 20.5 as
20.5x 0.
So, the degree of the
monomial is 0.
5. You can write the polynomial 4  9z in standard form as
9z + 4.
The greatest degree is 1, so the degree of the polynomial is 1.
The leading coefcient is 9.
The polynomial has 2 terms, so it is a binomial.
6. You can write the polynomial t 2  t 3  10 t in standard form
as t 3 + t 2  10 t.
The greatest degree is 3, so the degree of the polynomial is 3.
The leading coefcient is 1.
The polynomial has 3 terms, so it is a trinomial.
7. You can write the polynomial 2.8x + x 3 in standard form as
x 3 + 2.8x.
The greatest degree is 3, so the degree of the polynomial is 3.
The leading coefcient is 1.
The polynomial has 2 terms, so it is a binomial.
8. (b  10) + (4b  3) = b + 4b  10  3
= (b + 4b) + (10  3)
= 5b + (13)
= 5b  13
The sum is 5b  13.
380
 6) ⇒
3k 2  k + 11
2. In the monomial 7c 3d 2, the exponent of c is 3, and the
exponent of d is 2. So, the degree of the monomial is
3 + 2, or 5.
k+5
3k 2
Algebra 1
Worked-Out Solutions
The polynomial 25t + 100 represents the distance
between the objects after t seconds.
b. When t = 0, the distance between the objects is
25(0) + 100 = 100 feet. So, the constant term 100
indicates the distance between the penny and the
paintbrush is 100 feet when they begin to fall.
As the value of t increases by 1, the value of 25t + 100
decreases by 25. This means that the objects become
25 feet closer to each other after each second. So, the
coefcient 25 of the linear term represents how much
the distance between the objects changes each second.
7.1 Exercises (pp. 362–364)
Vocabulary and Core Concept Check
1. A polynomial in one variable is in standard form when the
exponents of the terms decrease from left to right.
2. The polynomial must have three terms, and the highest
degree must be 5. Sample answer: One possible polynomial
is 2x 5  x 2 + 7.
3. To determine whether a set of numbers is closed under an
operation, determine if the operation performed on any two
numbers in the set results in a number that is also in the set.
4. The expression that does not belong with the other three is
x 2  8 x. This expression has a variable in one of its
exponents, so it is not a polynomial. The other three
expressions are polynomials because the terms are constants
or other monomials whose variables have only whole
number exponents.
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Chapter 7
Monitoring Progress and Modeling with Mathematics
5. In the monomial 4g, the exponent of g is 1. So, the degree of
the monomial is 1.
form as 3z 4 + 2z 3 + 5z.
The greatest degree is 4, so the degree of the polynomial is 4.
6. In the monomial 23x 4, the exponent of x is 4. So, the degree
of the monomial is 4.
7. In the monomial 1.75k 2, the exponent of k is 2. So, the
degree of the monomial is 2.
4
18. You can write the polynomial 5z + 2z 3 + 3z 4 in standard
4
8. You can rewrite —9 as —9 x 0. So, the degree of the monomial
is 0.
9. In the monomial 7s 8t, the exponent of s is 8, and the exponent
of t is 1. So, the degree of the monomial is 8 + 1, or 9.
10. In the monomial 8m2n4, the exponent of m is 2, and the exponent
of n is 4. So, the degree of the monomial is 2 + 4, or 6.
11. In the monomial 9x y 3z7, the exponent of x is 1, the exponent
of y is 3, and the exponent of z is 7. So, the degree of the
monomial is 1 + 3 + 7, or 11.
12. In the monomial 3q 4rs6, the exponent of q is 4, the
exponent of r is 1, and the exponent of s is 6. So, the degree
of the monomial is 4 + 1 + 6, or 11.
13. You can write the polynomial 6c 2 + 2c 4  c in standard
form as 2c 4 + 6c 2  c.
The greatest degree is 4, so the degree of the polynomial is 4.
The leading coefcient is 2.
The leading coefcient is 3.
The polynomial has 3 terms, so it is a trinomial.
5
19. You can write the polynomial πr 2  —7 r 8 + 2r 5 in standard
5
form as —7 r 8 + 2r 5 + πr 2.
The greatest degree is 8, so the degree of the polynomial is 8.
5
The leading coefcient is —7 .
The polynomial has 3 terms, so it is a trinomial.
—
20. The polynomial √ 7 n 4 is in standard form.
The only term has a degree of 4, so the degree of the
polynomial is 4.
—
The leading coefcient is √7 .
The polynomial has 1 term, so it is a monomial.
4
21. The expression —3 πr 3 is a monomial because it is the product
of a number, —43 π, and a variable with a whole number
exponent, r 3. The only variable has an exponent of 3, so the
degree of the monomial is 3.
22. The expression 400x 8 + 600x 6 has two terms, so it is a
binomial. The greatest degree is 8, so the degree of the
binomial is 8.
23. (5y + 4) + (2y + 6) = 5y  2y + 4 + 6
= (5y  2y) + (4 + 6)
The polynomial has 3 terms, so it is a trinomial.
14. You can write the polynomial 4w11  w12 in standard form
as w12 + 4w11.
= 3y + 10
24. (8x  12) + (9x + 4) = 8x + 9x  12 + 4
The greatest degree is 12, so the degree of the polynomial is 12.
= (8x + 9x) + (12 + 4)
The leading coefcient is 1.
= x + (8)
=x8
The polynomial has 2 terms, so it is a binomial.
15. You can write the polynomial 7 + 3p 2 in standard form as
3p 2 + 7.
The greatest degree is 2, so the degree of the polynomial is 2.
The leading coefcient is 3.
The polynomial has 2 terms, so it is a binomial.
16. You can write the polynomial 8d  2  4d 3 in standard form
as
4d 3
+ 8d  2.
The greatest degree is 3, so the degree of the polynomial 3.
Alternate solution:
8x  12
+ 9x + 4
x 8
25. ( 2n2  5n  6 ) + ( n2  3n + 11 )
= 2n2  n2  5n  3n  6 + 11
= ( 2n2  n2 ) + (5n  3n) + (6 + 11)
= n2  8n + 5
The leading coefcient is 4.
The polynomial has 3 terms, so it is a trinomial.
17. The polynomial 3t 8 is in standard form.
The only term has a degree of 8, so the degree of the
polynomial is 8.
The leading coefcient is 3.
The polynomial has 1 term, so it is a monomial.
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Algebra 1
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381
Chapter 7
( 3p 3 + 5p 2  2p ) + ( p 3  8p 2  15p )
26.
= 3p 3  p 3 + 5p 2  8p 2  2p  15p
= y 2  4y + 9  3y 2 + 6y + 9
) + (  ) + (2p  15p)
= 4p 3 + ( 3p 2 ) + (17p)
= ( y 2  3y 2 ) + (4y + 6y) + (9 + 9)
=(
=
3p 3
4p 3


p3
5p 2
8p 2
 17p
3p 2
= 4m2  m + 2 + 3m2  10m  4
3p 3 + 5p 2  2p
= ( 4m2 + 3m2 ) + (m  10m) + (2  4)
+ p 3  8p 2  15p
= 7m2  11m  2
4p 3  3p 2  17p
(
3g2
 g) + (
3g2
= 2y 2 + 2y + 18
34. ( 4m2  m + 2 )  ( 3m2 + 10m + 4 )
Alternate solution:
27.
33. ( y 2  4y + 9 )  ( 3y 2  6y  9 )
Alternate solution:
 8g + 4 )
4m2 
m+2
4m2 
= ( 3g2 + 3g2 ) + (g  8g) + 4
= 6g2 + (9g) + 4
28.
7m2  11m  2
35. ( k 3  7k + 2 )  ( k 2  12 ) = k 3  7k + 2  k 2 + 12
= 6g2  9g + 4
( 9r 2
= k 3  k 2  7k + (2 + 12)
+ 4r  7 ) +
( 3r 2
 3r )
= 9r 2 + 3r 2 + 4r  3r  7
= ( 9r 2 + 3r 2 ) + (4r  3r)  7
= 12r 2 + r  7
= k 3  k 2  7k + 14
36. (r  10)  ( 4r 3 + r 2 + 7r )
= r  10 + 4r 3  r 2  7r
= 4r 3  r 2 + (r  7r)  10
Alternate solution:
= 4r 3  r 2  8r  10
9r 2 + 4r  7
Alternate solution:
+ 3r 2  3r
r  10
12r 2 + r  7

( 4r 3
+
r2
r  10
+ 7r ) ⇒ +
29. ( 4a  a3  3 ) + ( 2a3  5a2 + 8 )
= a3 + 2a3  5a2 + 4a  3 + 8
= ( a3 + 2a3 )  5a2 + 4a (3 + 8)
= a3  5a2 + 4a + 5
30.
( s3
 2s  9 ) +
( 2s 2
=

6s 3 )
+
2s 2

+ s)
+ (2s + s)  9
= 5s 3 + 2s 2  s  9
+
6s 3
+
 7r
37. ( t 4  t 2 + t )  ( 12  9t 2  7t )
= t 4  t 2 + t  12 + 9t 2 + 7t
= t 4 + 8t 2 + 8t  12
Alternate solution:
t4  t2 + t
t4  t2 + t
 ( 9t 2  7t + 12 ) ⇒ + + 9t 2 + 7t  12
t 4 + 8t 2 + 8t  12
 2s  9
2s 2

r2
4r 3  r 2  8r  10
Alternate solution:
s3
4r 3
= t 4 + ( t 2 + 9t 2 ) + (t + 7t)  12
6s 3
= s 3  6s 3 + 2s 2  2s + s  9
( s3
m+2
 ( 3m2 + 10m + 4 ) ⇒ 3m2  10m  4
= 3g2 + 3g2  g  8g + 4
+s
5s 3 + 2s 2  s  9
31. (d  9)  (3d  1) = d  9  3d + 1
= (d  3d) + (9 + 1)
= 2d  8
32. (6x + 9)  (7x + 1) = 6x + 9  7x  1
= (6x  7x) + (9  1)
= x + 8
38. ( 4d  6d 3 + 3d 2 )  (10d 3 + 7d  2)
= 4d  6d 3 + 3d 2  10d 3  7d + 2
= ( 6d 3  10d 3 ) + 3d 2 + (4d  7d) + 2
= 16d 3 + 3d 2  3d + 2
39. When writing the subtraction as addition, the last term of the
polynomial was not multiplied by 1.
( x2 + x )  ( 2x2  3x ) = x2 + x  2x2 + 3x
= ( x2  2x2 ) + (x + 3x)
= x2 + 4x
Alternate solution:
6x + 9
6x + 9
 (7x + 1) ⇒ 7x  1
x + 8
382
Algebra 1
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Chapter 7
40. The terms 4x2 and 8x are not like terms, so they cannot be
added.
x3  4x2
+ 3x3
51. 16t 2 + v0t + s0 = 16t 2 + (45)t + 200
= 16t 2  45t + 200
+3
Let t = 1.
+ 8x  2
16t 2  45t + 200 = 16(1)2  45(1) + 200
2x3  4x2 + 8x + 1
= 16  45 + 200
= 139
41. (8b + 6)  (4 + 5b) = 8b + 6  4  5b
= (8b  5b) + (6  4)
= 3b + 2
A polynomial that represents how much more it costs to
make b necklaces than b bracelets is 3b + 2.
42. (142 + 12m) + (52 + 6m) = 142 + 52 + 12m + 6m
= (142 + 52) + (12m + 6m)
A polynomial that represents the height of the water balloon
is 16t 2  45t + 200, and the water balloon is 139 feet
from the ground after 1 second.
52. 16t 2 + v0t + s0 = 16t 2 + 16t + 3
Let t = 1.
16t 2 + 16t + 3 = 16(1)2 + 16(1) + 3
= 16(1) + 16 + 3
= 194 + 18m
A polynomial that represents the total number of
memberships at the tness center is 194 + 18m.
2s2  5st  t 2
43.

( s2
+ 7st 
t2 )
2s2  5st  t 2
⇒ + s2  7st + t 2
s2  12st
44. ( a2  3ab + 2b2 ) + ( 4a2 + 5ab  b2
)
= 16 + 16 + 3
=3
The polynomial 16t 2 + 16t + 3 represents the height of
the tennis ball after t seconds, and the tennis ball is 3 feet
high after 1 second. So, it is back to the initial height where
you hit it with the racket.
53. a. ( 16t 2 + 98 )  ( 16t 2 + 46t + 6 )
= 16t 2 + 98 + 16t 2  46t  6
= a2  4a2  3ab + 5ab + 2b2  b2
= ( 16t 2 + 16t 2 )  46t + (98  6)
= ( a2  4a2 ) + (3ab + 5ab) + ( 2b2  b2 )
= 46t + 92
= 3a2 + 2ab + b2
45.
 6d 2
The polynomial 46t + 92 represents the distance
between the objects after t seconds.
+ c 2  2cd + 2d 2
b. When t = 0, the distance between the objects is
c2
46(0) + 92 = 92 feet. So, the constant term 92
indicates that the distance between the two balls is
92 feet when they are thrown.
2c 2  2cd  4d 2
46. ( x 2 + 9xy )  ( x 2 + 6xy  8y 2 )
= x 2 + 9xy  x 2  6xy + 8y 2
As the value of t increases by 1, the value of 46t + 92
decreases by 46. This means that the two balls become
46 feet closer to each other after each second. So, the
coefcient, 46, of the linear term represents how much
the distance between objects changes each second.
= ( x2  x2 ) + (9xy  6xy) + 8y 2
= 2x 2 + 3xy + 8y 2
47. The terms of a polynomial are always monomials.
A polynomial is a monomial or a sum of monomials, and
each monomial is a term of the polynomial.
48. The difference of two trinomials is sometimes a trinomial. If
like terms have the same coefcient, they will cancel when
subtracted, so the difference will have fewer than
3 terms. Or, if the terms in the trinomial are not all of the same
degree, then the difference could have more than 3 terms.
49. A binomial is sometimes a polynomial of degree 2. The two
terms in the binomial can be of any degree.
50. The sum of two polynomials is always a polynomial.
Polynomials are closed under addition.
0.028t 3 + 0.06t 2 + 0.1t + 17
54. a.
+
 0.38t 2 + 1.5t + 42
0.028t 3  0.32t 2 + 1.6t + 59
The polynomial 0.028t 3  0.32t 2 + 1.6t + 59
represents the total amount spent each year on buying new
and used vehicles in the 7-year period.
b. Let t = 5.
0.028t 3  0.32t 2 + 1.6t + 59
= 0.028(5)3  0.32(5)2 + 1.6(5) + 59
= 0.028(125)  0.32(25) + 8 + 59
= 3.5  8 + 8 + 59
= 55.5
The total amount spent on buying new and used vehicles
in the fth year was $55.5 million.
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Algebra 1
Worked-Out Solutions
383
Chapter 7
b. 2x2  14x = 2(20)2  14(20)
55. P = 2x + (3x  2) + (2x + 1) + (5x  2)
= 2(400)  280
= (2x + 3x + 2x + 5x) + (2 + 1  2)
= 800  280
= 12x  3
= 520
The polynomial 12x  3 represents the perimeter of the
quadrilateral.
56. a. 2x + 1  (x  2)
x  2  (2x + 1)
or
= 2x + 1  x + 2
= x  2 + 2x  1
= (2x  x) + (1 + 2)
= (x + 2x) + (2  1)
= 3x + 3
3
0
3
+3 +3
3x = 3
3x 3
—=—
3
3
x=1
3x = 3
3x 3
—=—
3
3
x=1
a sum of
1
—2 x 2 +
+ —12 x 2 

(
1
—2 x 2
1
—2 x 2
+

1
—2
1
—2
)
Area of
Area of
Area of
Area of
patio = rectangular area  pool  hot tub
=
ℓw

ℓw

s2
= 8x(4x)  10(4x)  (12x2  28x)  x2
= 32x2  40x  12x2 + 28x  x2
= ( 32x2  12x2  x2 ) + (40x + 28x)
= 19x2  12x
b. 19x2  12x = 19(9)2  12(9)
1
—2
+ and
1
—2 x 2
1
—2
 have
and a difference of 1.
1
—2
1
—2
61. a.
= (8x  10)(4x)  ( 6x(2x)  14(2x) )  x2
associative, so you can add in any order.
x2
So, you need 1.3 gallons of sealant to cover the deck once.
A = (8x  10)(2x + 2x)  (6x  14)(2x)  x2
57. Your friend is correct. Addition is commutative and
1
—2 x 2
= 1.3
—
400
A
Each expression equals 0 when x = 1, which is the
x-coordinate of the point of intersection of the two lines.
58. Sample answer: The polynomials
covered by 1 gallon of sealant to get the amount of sealant
needed.
520
3x  3 = 0
or
c. Divide the total area of the deck by the area that can be
= 3x  3
or
The vertical distance between points on the lines with the
same x-value can be represented by the absolute value of
3x + 3 or 3x  3.
b. 3x + 3 =
When x = 20 feet, the total area of the deck is
520 square feet.
= 
1
—2 x 2
1
—2 x 2
+
+
x2
1
—2
1
—2
1
59. a. The set of negative integers is not closed under
multiplication because the product of two negative
integers is always a positive integer.
b. The set of whole is closed under addition because the sum
of two whole numbers is always a whole number.
= 19(81)  108
= 1539  108
= 1431
So, when x = 9, the area of the patio is 1431 square feet.
The patio costs 1431(10) = $14,310.
Maintaining Mathematical Prociency
62. 2(x  1) + 3(x + 2) = 2(x)  2(1) + 3(x) + 3(2)
= 2x  2 + 3x + 6
= (2x + 3x) + (2 + 6)
= 5x + 4
63. 8(4y  3) + 2( y  5) = 8(4y)  8(3) + 2( y)  2(5)
60. a. Level 1: A = ℓ w
= x( 10 + (x  12) )
= x( x + (10  12) )
= x(x  2)
= x(x)  x(2)
= x 2  2x
Level 2: A = ℓ w
= x(x  12)
= x(x)  x(12)
= x 2  12x
= 32y  24 + 2y  10
= (32y + 2y) + (24  10)
= 34y  34
64. 5(2r + 1)  3(4r + 2) = 5(2r) + 5(1)  3(4r)  3(2)
= 10r + 5 + 12r  6
= (10r + 12r) + (5  6)
= 22r  1
Total area:
( x 2  2x ) + ( x 2  12x ) = x 2 + x 2  2x  12x
= ( x 2 + x 2 ) + (2x  12x)
= 2x 2  14x
384
Algebra 1
Worked-Out Solutions
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Chapter 7
7.2 Explorations (p. 365)
7.2 Monitoring Progress (pp. 366–368)
1. a. 1; The product of 1 and 1 is 1.
1. ( y + 4)( y + 1) = y( y + 1) + 4( y + 1)
b. 1; The product of 1 and 1 is 1.
= y( y) + y(1) + 4( y) + 4(1)
c. 1; The product of 1 and 1 is 1.
= y 2 + y + 4y + 4
d. x; The product of any number and 1 is that number.
= y 2 + 5y + 4
e. x; The product of any number and 1 is the opposite of
that number.
f. x; The product of any number and 1 is that number.
The product is y2 + 5y + 4.
×
g. x; The product of the opposite of a number and 1 is the
i.
z+6
6z  12
number.
h.
z2
2.
The product of any number multiplied by itself is the
number squared.
z2  2z
x2;
The product is z2 + 4z  12.
x2;
The product of any number and its opposite is the
opposite of the number squared.
j. x 2; The product of any number multiplied by itself is the
z2 + 4z  12
3. ( p + 3)( p  8) = ( p + 3)[ p + (8) ]
number squared.
2. a. (x + 3)(x  2) =
x2
+ (2x + 3x)  6
=
x2
+x6
p
8
b. (2x  1)(2x + 1) = 4x 2 + (2x  2x)  1
=
4x2
1
c.
(x + 2)(2x  1) =
2x2
d.
3
3p
8p
24
The product is
p2
+ 3p  8p  24, or p2  5p  24.
4. (r  5)(2r  1) = [ r + (5) ][ 2r + (1) ]
+ (x + 4x)  2
= 2x2 + 3x  2
p
p2
r
5
2r
2r2
10r
1
r
5
The product is 2r 2  10r  r + 5, or 2r 2  11r + 5.
First
5.
Outer
Inner
Last
(m  3)(m  7) = m(m) + m(7) + (3)(m) + (3)(7)
= m2 + (7m) + (3m) + 21
= m2  10m + 21
The product is m2  10m + 21.
(x  2)(x  3) = x2 + (3x  2x) + 6
First Outer
6.
= x2 + x + 6
Inner
(x  4)(x + 2) = x(x) + x(2) + (4)(x) + (4)(2)
= x2 + 2x + (4x) + (8)
3. Multiply each term in one polynomial by each term in the
= x2  2x  8
other polynomial, then combine like terms.
The product is x2  2x  8.
4. Sample answer:
First
7.
Outer
Inner
( 2u + )( u  ) = 2u(u) + 2u ( )
1
—2
3
—2
3
—2
= 2u2 + (3u) +
(x + 1)(x  2) = x2 + (2x + x)  2
= x2  x  2
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Last
+ —12 (u)
1
3
—2 u  —4
Last
+
1
—2
( — )
3
2
= 2u2  —52 u  —34
The product is 2u2  —52 u  —34 .
Algebra 1
Worked-Out Solutions
385
Chapter 7
First
8.
Outer
Inner
Last
(n + 2)( n2 + 3 ) = n( n2 )+ n(3) + 2( n2 ) + 2(3)
4. ( y + 6)( y + 4) = y( y + 4) + 6( y + 4)
= y( y) + y(4) + 6( y) + 6(4)
= n3 + 3n + 2n2 + 6
=
n3
+
2n2
= y2 + 4y + 6y + 24
+ 3n + 6
= y2 + 10y + 24
The product is n3 + 2n2 + 3n + 6.
5. (z  5)(z + 3) = z(z + 3)  5(z + 3)
x2 + 5x + 8
9.
×
= z(z) + z(3)  5(z)  5(3)
x+1
= z2 + 3z  5z  15
x2 + 5x + 8
x3
+
5x2
= z2  2z  15
+ 8x
6. (a + 8)(a  3) = a(a  3)+ 8(a  3)
x3 + 6x2 + 13x + 8
= a(a)  a(3) + 8(a)  8(3)
The product is x3 + 6x2 + 13x + 8.
n2
10.
= a2  3a + 8a  24
 2n + 4
×
= a2 + 5a  24
n3
7. (g  7)(g  2) = g(g  2)  7(g  2)
3n2 + 6n  12
= g(g)  g(2)  7(g)  7(2)
n3  2n2 + 4n
= g2  2g  7g + 14
n3  5n2 + 10n  12
= g2  9g + 14
The product is n3  5n2 + 10n  12.
8. (n  6)(n  4) = n(n  4)  6(n  4)
1
11. —2 h ( b1 + b2 ) = —2 (x  7)[ x(x + 11) ]
1
= n(n)  n(4)  6(n)  6(4)
= —12 (x  7)(2x + 11)
F
O
I
= n2  4n  6n + 24
L
= —12 [ 2x2 + 11x + (14x) + (77) ]
= —12 ( 2x2  3x  77 )
= n2  10n + 24
9. (3m + 1)(m + 9) = 3m(m + 9) + 1(m + 9)
= 3m(m) + 3m(9) + 1(m) + 1(9)
77
= x2  —32 x  —
2
The polynomial that represents the area of the trapezoidal
77
region becomes x2  —32 x  —
. The longer base becomes
2
x + 11. Substituting this value in the formula for the area of
a trapezoid along with the other unchanged values changes
the linear and constant terms in the polynomial.
= 3m2 + 27m + m + 9
= 3m2 + 28m + 9
10. (5s + 6)(s  2) = 5s(s  2) + 6(s  2)
= 5s(s)  5s(2) + 6(s)  6(2)
= 5s2  10s + 6s  12
= 5s2  4s  12
7.2 Exercises (pp. 369–370)
Vocabulary and Core Concept Check
11. (x + 3)(x + 2)
1. Sample answer: Distribute one of the binomials over each
term in the other binomial, and simplify by combining like
terms. Or, write each binomial as a sum of terms and make
a table of products. Then, write the sum of the products and
simplify by combining like terms.
2. The letters stand for the sets of terms to multiply: rst, outer,
x
3
x
x2
3x
2
2x
6
x2 + 3x + 2x + 6 = x2 + 5x + 6
12. ( y + 10)( y  5) = ( y + 10)[ y + (5) ]
inner, and last.
Monitoring Progress and Modeling with Mathematics
3. (x + 1)(x + 3) = x(x + 3) + 1(x + 3)
= x(x) + x(3) + 1(x) + 1(3)
= x2 + 3x + x + 3
= x2 + 4x + 3
386
Algebra 1
Worked-Out Solutions
y
5
y
10
y2
10y
5y 50
y2 + 10y  5y  50 = y2 + 5y  50
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Chapter 7
13. (h  8)(h  9) = [ h + (8) ][ h + (9) ]
h
9
h
8
h2
8h
9h
72
20. The two terms that represent x  5 on the left side of the
table should be x and 5, not 5.
(x  5)(3x + 1) = [ x + (5) ](3x + 1)
x
h2  8h  9h + 72 = h2  17h + 72
c
5
6
c2
6c
5c
30
1
3x 2
x
5 15x 5
14. (c  6)(c  5) = [ c + (6) ][ c + (5) ]
c
3x
3x2 + x  15x  5 = 3x2  14x  5
First
21.
= b2 + 7b + 3b + 21
= b2 + 10b + 21
c2  6c  5c + 30 = c2  11c + 30
15. (3k  1)(4k + 9) = [ 3k + (1) ](4k + 9)
3k
1
4k
12k2
4k
9
27k
9
First
22.
g
5g2
3g
8
40g
24
5g2 +
3g + 40g + 24 =
4j
3
8j 2
12j
7 14j
First
23.
Last
= k2 + 4k  5
First Outer
24.
Inner
Last
(x  4)(x + 8) = x(x) + x(8) + (4)(x) + (4)(8)
5g2
= x2 + 8x  4x  32
+ 43g + 24
= x2 + 4x  32
First Outer
25.
1
—4
3
—4
First
26.
3
4
1
4
Outer
Inner
Last
( z  — )( z  — ) = z(z) + z( — ) + ( — )(z) + ( — )( — )
5
3
2
3
2
3
5
3
5
3
2
3
10
= z2  —23 z  —53 z + —
9
84
10
= z2  —73 z + —
9
15d 2  36d  35d + 84 = 15d 2  71d + 84
27.
First
Outer
Inner
Last
(9  r)(2  3r) = 9(2) + 9(3r) + (r)(2) + (r)(3r)
= t(t) + t(5)  2(t)  2(5)
= 18  27r  2r + 3r 2
= t2 + 5t  2t  10
= 18  29r + 3r 2
= t2 + 3t  10
= 3r 2  29r + 18
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Last
3
= q2  —12 q  —
16
12
(t  2)(t + 5) = t (t + 5)  2(t + 5)
1
—4
3
= q2 + —14 q  —34 q  —
16
21
19. The rst term t also should be multiplied by t + 5.
Inner
( q  )( q + ) = q(q) + q( ) + ( ) (q) + ( — )( — )
3
—4
15d 2 36d
7 35d
Inner
= k2  k + 5k  5
18. (5d  12)(7 + 3d) = [ 5d + (12) ][ 3d + (7) ]
3d
Outer
(k + 5)(k  1) = k(k) + k(1) + 5(k) + 5(1)
8j2  12j  14j + 21 = 8j 2  26j + 21
5d
Last
= w2 + 15w + 54
17. (3 + 2j )(4j  7) = [ 2j + (3) ][ 4j + (7) ]
2j
Inner
= w2 + 6w + 9w + 54
16. (5g + 3)(g + 8)
3
Outer
(w + 9)(w + 6) = w(w) + w(6) + 9(w) + 9(6)
12k2  4k + 27k  9 = 12k2 + 23k  9
5g
Outer Inner Last
(b + 3)(b + 7) = b(b) + b(7) + 3(b) + 3(7)
Algebra 1
Worked-Out Solutions
387
Chapter 7
First
28.
Outer
Inner
Last
33.
(8  4x)(2x + 6) = 8(2x) + 8(6) + (4x)(2x) + (4x)(6)
= 16x + 48  8x2  24x
Area of
Area of
Area of
=

shaded region
rectangular region
triangular region
A
= 8x2 + 16x  24x + 48
= 8x2  8x + 48
First
29.
Outer
Inner
F
Last
(w + 5)( w2 + 3w ) = w( w2 ) + w(3w) + 5( w2 ) + 5(3w)
Outer
Inner
Last
(
=
L
= 2x(x) + 2x(5) + (9)(x) + (9)(5)
34.
+ 10x  9x  45
+
11
x
—
2
 ℓw
 (x  7)(5)
1)2
= (x + 1)(x + 1)  (x  7)(5)
F
O
I
L
= x(x) + x(1) + 1(x) + 1(1)  [ x(5)  7(5) ]
= —12 (p + 1)(2p  6)
=
+ 5x + 6x + 30 )
)
= (x +
1
=
[ x(x) + x(5) + 6(x) + 6(5) ]
= s2
A
32. A = —2 bh
=
L
+ 15
The polynomial
+ x  45 represents the area of the
rectangular region.
1
=—2
I
+ 11x + 30 )
) (
1
—2 x 2
2x2
O
1
—2 ( x 2
1
—2 ( x 2
O
Area of
Area of
Area of
=

shaded region
square region
rectangular region
= 2x2 + x  45
F
F
 —21
11
The polynomial —12 x2 + —
x + 15 represents the area of the
2
shaded region.
= (2x  9)(x + 5)
=
L
11
= x2  —12 x2 + 11x  —
x + (30  15)
2
31. A = ℓ w
2x2
I
11
= x2 + 11x + 30  —12 x2 + —
x  15
2
= v3 + 5v2  24v
I
O
= x2 + 11x + 30  —12 ( x2 )  —12 (11x)  —12 (30)
= v3 + 8v2  3v3  24v
O
 —12 (x + 6)(x + 5)
= (x2 + 11x + 30) 
(v  3)( v2 + 8v ) = v( v2 ) + v(8v) + (3)( v2 ) + (3)(8v)
F
= (x + 6)(x + 5)
= ( x2 + 5x + 6x + 30 ) 
= w3 + 8w2 + 15w
First
 —12 bh
= [ x(x) + x(5) + 6(x) + 6(5) ]
= w3 + 3w2 + 5w2 + 15w
30.
= ℓw
I
= x2 + x + x + 1  (5x  35)
L
[ p(2p) + p(6) + 1(2p) + 1(6) ]
( 2p2  6p + 2p  6 )
( 2p2  4p  6 )
( 2p2 )  —12 (4p)  —12 (6)
1
—2
1
—2
1
—2
= p2  2p  3
= x2 + 2x + 1  5x + 35
= x2 + 2x  5x + 1 + 35
= x2  3x + 36
The polynomial x2  3x + 36 represents the area of the
shaded region.
x2 + 3x + 2
35.
×
The polynomial p2  2p  3 represents the area of the
triangular region.
x+4
4x2 + 12x + 8
x3 + 3x2 + 2x
x3 + 7x2 + 14x + 8
f 2 + 4f + 8
36.
×
f+1
f 2 + 4f + 8
f 3 + 4f 2 + 8f
f 3 + 5f 2 + 12f + 8
y2 + 8y  2
37.
×
y+3
3y2 + 24y  6
y3 + 8y2  2y
y3 + 11y2 + 22y  6
388
Algebra 1
Worked-Out Solutions
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Chapter 7
t2  5t + 1
38.
×
44. a. A = ℓ w
t2
= (x + 22 + x)(x + 20 + x)
2t2 + 10t  2
t3

5t2
= (2x + 22)(2x + 20)
+t
F
t3  7t2 + 11t  2
×
A polynomial that represents the combined area of the
photo and the frame is ( 4x2 + 84x + 440 ) square inches.
20b2 + 20b  16
5b3  5b2 + 4b
b. Let x = 4.
5b3 + 15b2 + 24b  16
40.
4x2 + 84x + 440 = 4(4)2 + 84(4) + 440
= 4(16) + 336 + 440
d+7
×
= 64 + 336 + 440
d+6
2d 3

d2
= 840
 6d + 42
12d 2
When the width of the frame is 4 inches, the combined
area of the photo and the frame is 840 square inches.
+ 7d
2d 3 + 11d 2 + d + 42
45. The degree of the product is the sum of the degrees of each
3e2  5e + 7
41.
×
binomial.
6e + 1
46. Sample answer: (2x  6)( x 2 + 3x + 4 )
3e2  5e + 7
18e3  30e2 + 42e
x2 + 3x + 4
18e3  27e2 + 37e + 7
6v2
42.
×
2x  6
6x2
+ 2v  9
5v + 4

10v2
2x3
10x  24
The product is 2x3  10x  24, which is a trinomial of degree 3.
+ 45v
30v3 + 14v2 + 53v  36
47. The FOIL method can only be used for multiplying two
43. a. A = ℓ w
= (10x + 10)(4x + 20)
F
 18x  24
2x3 + 6x2 + 8x
24v2 + 8v  36
30v3
L
= 4x2 + 84x + 440
b + 4
2d2
I
= 4x2 + 40x + 44x + 440
5b2 + 5b  4
39.
O
= 2x(2x) + 2x(20) + 22(2x) + 22(20)
O
I
L
=10x(4x) + 10x(20) + 10(4x) + 10(20)
= 40x2 + 200x + 40x + 200
= 40x2 + 240x + 200
A polynomial that represents the area of the football eld
is ( 40x2 + 240x + 200 ) square feet.
binomials, because each of the four letters represent one of
the products when two binomials are multiplied. When two
trinomials are multiplied, however, there are 6 products. The
FOIL method would leave out the products that include the
middle terms of the two trinomials.
48. a. The two binomials being multiplied are (4x + 3) and
(8x  9). The binomial (4x + 3) is comprised of the
two terms on top of the table, and the binomial (8x  9)
is comprised of the two terms on the left side of the table.
b. When x > 0, a is positive because it is the product of two
b. 4x + 20 = 160
negative terms, b is negative because it is the product of one
negative term and one positive term, c is positive because it is
the product of two negative terms, and d is negative because
it is the product of one negative term and one positive term.
 20  20
4x = 140
4x 140
—=—
4
4
x = 35
49. Your answers should be equivalent. You are both multiplying
40x2 + 240x + 200 = 40(35)2 + 240(35) + 200
the same binomials, and neither the order in which you
multiply nor the method used will make a difference.
= 40(1225) + 8400 + 200
= 49,000 + 8400 + 200
= 57,600
When the width of a football eld is 160 feet, the area is
57,600 square feet.
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Algebra 1
Worked-Out Solutions
389
Chapter 7
54. y = 4 ∣ x + 2 ∣
50. V = ℓ wh
= (4x  3)(x + 1)(x + 2)
= (4x  3)[ x(x + 2) + 1(x + 2) ]
= (4x  3)[ x(x) + x(2) + 1(x) + 1(2) ]
= (4x 
3)( x2
+ 2x + x + 2 )
4[ (x + 2) ],
{4(x
+ 2),
if x + 2 < 0
if x + 2 ≥ 0
g(x) = 4[ (x + 2) ],
if x + 2 <
g(x) = 4(x  2),
2
= (4x  3)( x2 + 3x + 2 )
g(x) = 4(x)  4(2),
= 4x( x2 + 3x + 2 )  3( x2 + 3x + 2 )
g(x) = 4x + 8,
= 4x( x2 ) + 4x(3x) + 4x(2)  3( x2 )  3(3x)  3(2)
g(x) = 4(x + 2),
=
+
if x < 2
if x + 2 ≥
0
2
g(x) = 4x  8,
if x ≥ 2
= 4x3 + 9x2  x  6
So, a piecewise function for y = 4 ∣ x + 2 ∣ is
4x + 8,
if x < 2
.
g(x) =
4x  8, if x ≥ 2
3x2
 9x  6
2
g(x) = 4(x)  4(2),
12x2
+ 8x 
0
= 4x3 + ( 12x2  3x2 ) + (8x  9x)  6
4x3
A polynomial that represents the volume of the container is
( 4x3 + 9x2  x  6 ) cubic feet.
51. a. The product of m and n is c. So, when c > 0, m and n have
the same signs because the product of two numbers with
the same sign is positive.
b. The product of m and n is c. So, when c < 0, m and n have
opposite signs because the product of one positive number
and one negative number is negative.
Maintaining Mathematical Prociency
52. y = ∣ x ∣ + 4
A piecewise function for y = ∣ x ∣ + 4 is
g(x) =
x + 4, if x < 0
.
x + 4,
if x ≥ 0
{
g(x) =
6[ (x  3) ],
{ 6(x
 3),
g(x) = 6[ (x  3) ],
g(x) = 6(x + 3),
g(x) = 6(x  3),
g(x) = 6(x)  6(3)
g(x) = 6x  18,
{
55. 10 2
⋅ 10
9
⋅
= 10 2 + 9
= 1011
x5 + 1 x6
1
x
x
x
1
1
1
1
57. (3z 6)3 = —
=—=—=—
(3z 6)3 33(z 6)3 27z 6 ⋅3 27z18
x5 x
x
56. —
=—
= —8 = x 6  8 = x2 = —2
8
8
( )
2y 4
y
58. —
3
2
1
1
= (2y 4  3)2 = (2y)2 = 22y2 = —
=—
22y2 4y2
7.3 Explorations (p. 371)
1. a. x 2  2x + 2x  4 = x 2  4
2. a.
if x  3 < 0
if x  3 ≥ 0
if x  3 <
0
+3 +3
g(x) = 6(x) + 6(3),
g(x) = 6x + 18,
2
b. 4x 2 + 2x  2x  1 = 4x 2  1
53. y = 6 ∣ x  3 ∣
if x  3 ≥
+3
x 2 + 2x + 2x + 4 = x 2 + 4x + 4
b.
if x < 3
0
+3
if x ≥ 3
So, a piecewise function for y = 6 ∣ x  3 ∣ is
6x + 18, if x < 3
.
g(x) =
6x  18,
if x ≥ 3
{
390
g(x) =
Algebra 1
Worked-Out Solutions
4x2  2x  2x + 1 = 4x2  4x + 1
Copyright © Big Ideas Learning, LLC
All rights reserved.
Chapter 7
F
3.
O
I
L
(a + b)(a  b) = a(a) + a(b) + b(a) + b(b)
= a2  ab + ab  b 2
(3x + 1)(3x  1) = (3x)2  12
= 9x2  1
b2
+
+
+
−
+
+
+
+
+
+
+
+
+
+
+
+
−
Check
−

(a + b)(a  b) = a2  b2
−
=
a2
c.
+
+
+
+
−
So, (a + b)(a  b) = a2  b 2.
(a + b)2 = (a + b)(a + b)
F
O
I
L
= a(a) + a(b) + b(a) + b(b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
So, (a + b)2 = a2 + 2ab + b2.
(a  b)2 = (a  b)(a  b)
F
O
I
L
= a(a) +a(b) + (b)(a) + (b)(b)
=
a2
 ab  ab +
So, (a 
=
a2
(x + 3)2 = x2 + 2(x)(3) + 32
 2ab +
= x2 + 6x + 9
b2.
Check
+
+ + +
+
+
+ + +
− − −
+
+
+
+
+ + +
+ + +
+
+
+ + +
4. a. ( a + b)( a  b) = a2  b 2
(x + 3)(x  3) =
x2
The product is 9x2  1.
d. (a + b)2 = a2 + 2ab + b 2
b2
= a2  2ab + b2
b)2
9x2  3x + 3x  1 = 9x2  1

32
= x2  9
Check
+
+
−
−
+
+
+
+
+
+
− − −
− − −
− − −
−
+
x2 + 3x + 3x + 9 = x2 + 6x + 9
The product is x2 + 6x + 9.
e. (a  b)2 = a2  2ab + b 2
(x  2)2 = x2  2(x)(2) + 22
x2  3x + 3x  9 = x2  9
= x2  4x + 4
The product is x2  9.

− −
+
−
−
+ +
−
−
+ +
42
= x2  16
Check
+
−
−
(x  4)(x + 4) =
x2
+
Check
b. (a  b)(a + b) = a2  b 2
+
+ + + +
+
+
+ + + +
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
x2  2x  2x + 4 = x2  4x + 4
The product is x2  4x + 4.
x2 + 4x  4x  16 = x2  16
The product is x2  16.
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Algebra 1
Worked-Out Solutions
391
Chapter 7
9. a. The Punnett square shows four possible gene
f. (a + b)2 = a2 + 2ab + b 2
combinations of the offspring. Of these combinations,
one results in black. So, —14 = 25% of the possible gene
combinations result in black.
(3x + 1)2 = (3x)2 + 2(3x)(1) + 12
= 9x2 + 6x + 1
Check
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Expand the product to nd the possible colors of the
offspring.
+
+
+
+
+
(0.5B + 0.5W )2 = (0.5B)2 + 2(0.5B)(0.5W ) + (0.5W )2
9x2 + 3x + 3x + 1 = 9x2 + 6x + 1
The product is 9x2 + 6x + 1.
7.3 Monitoring Progress (pp. 372–374)
1. (x + 7)2 = x 2 + 2(x)(7) + 72
=
x2
+ 14x + 49
The product is x2 + 14x + 49.
2. (7x  3)2 = (7x)2  2(7x)(3) + 32
= 49x2  42x + 9
The product is 49x2  42x + 9.
3. (4x  y)2 = (4x)2  2(4x)(y) + y2
= 16x2  8xy + y2
The product is 16x2  8xy + y2.
4. (3m + n)2 = (3m)2 + 2(3m)(n) + n2
= 9m2 + 6mn + n2
The product is 9m2 + 6mn + n2.
5. (x + 10)(x  10) = x 2  102
= x2  100
The product is x2  100.
6. (2x + 1)(2x  1) = (2x)2  12
= 4x2  1
The product is 4x2  1.
7. (x + 3y)(x  3y) = x 2  (3y)2
= x2  9y2
The product is x2  9y2.
8. Rewrite 212 as (20 + 1)2. Then use the square of a binomial
pattern.
(20 + 1)2 = 202 + 2(20)(1) + 12
= 400 + 40 + 1
= 441
b. Model the gene from each parent with the expression
0.5B + 0.5W. There is an equal chance that the offspring
inherits a black or a white gene from each parent.
You can model the possible gene combinations of the
offspring with the expression (0.5B + 0.5W )2. Notice that
this product also represents the area of the Punnett square.
= 0.25B2 + 0.5BW + 0.25W 2
Consider the coefcients in the polynomial. So, 25% of
the offspring are BB, or black; 50% of the offspring are
BW, or gray; and 25% of the offspring are WW, or white.
7.3 Exercises (pp. 375–376)
Vocabulary and Core Concept Check
1. Substitute the rst term of the binomial for a and the second
term of the binomial for b in the square of a binomial pattern
a2 + 2ab + b2. Then simplify.
2. The expression that does not belong is (x + 2)(x  3). It
is the only one that cannot be simplied using the sum and
difference pattern. The pattern for this one is (a + b)(a  c),
but the pattern for the others is (a + b)(a  b).
Monitoring Progress and Modeling with Mathematics
3. (x + 8)2 = x 2 + 2(x)(8) + 82
= x2 + 16x + 64
4. (a  6)2 = a2  2(a)(6) + 62
= a2  12a + 36
5. (2f  1)2 = (2f )2  2(2f )(1) + 12
= 4f 2  4f + 1
6. (5p + 2)2 = (5p)2 + 2(5p)(2) + 22
= 25p2 + 20p + 4
7. (7t + 4)2 = (7t)2 + 2(7t)(4) + 42
= 49t2  56t + 16
8. (12  n)2 = (12)2  2(12)(n) + n2
= 144 + 24n + n2
= n2 + 24n + 144
9. (2a + b)2 = (2a)2 + 2(2a)(b) + b 2
= 4a2 + 4ab + b2
10. (6x  3y)2 = (6x)2  2(6x)(3y) + (3y)2
= 36x2  36xy + 9y2
392
Algebra 1
Worked-Out Solutions
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Chapter 7
11. (x + 4)2 = x 2 + 2(x)(4) + 42
=
x2
27. 422 = (40 + 2)2
+ 8x + 16
= 402 + 2(40)(2) + 22
= 1600 + 160 + 4
12. (x + 7 + x)2 = (2x + 7)2
=
+ 2(2x)(7) +
(2x)2
= 1764
72
= 4x 2 + 28x + 49
28. 292 = (30  1)2
= 302  2(30)(1) + 12
13. (7n  5)2 = (7n)2  2(7n)(5) + 52
= 900  60 + 1
= 49n2  70n + 25
14. (4c + 4d )2 = (4c)2 + 2(4c)(4d ) + (4d )2
= 16c2 + 32cd + 16d 2
= 841
29. 30.52 = (30 + 0.5)2
= 302 + 2(30)(0.5) + 0.52
15. (t  7)(t + 7) = t 2  72
= 900 + 30 + 0.25
= t2  49
= 930.25
16. (m + 6)(m  6) =
m2

=
m2
 36
17. (4x + 1)(4x  1) =
62
(4x)2

1
30. 10 —3
⋅ 9 — = ( 10 + — ) ( 10  — )
2
3
= 102 
12
= 100 
= 16x2  1
18. (2k  4)(2k + 4) =
=
19. (8 + 3a)(8  3a) =
(2k)2
4k2
82

(
1
)(
1
2
) ()
1
31. The middle term in the square of a binomial pattern was not
included.
(k + 4)2 = k 2 + 2(k)(4) + 42
(3a)2
= k 2 + 8k + 16
9a2
32. There is no middle term in the sum and difference pattern.
 c2
(s + 5)(s  5) = s 2  52
= —14  c2
= s 2  25
21. ( p  10q)( p + 10q) = p2  (10q)2
= p2  100q2
22. (7m + 8n)(7m  8n) = (7m)2  (8n)2
=
23. (y + 4)(y  4) =
49m2
(y)2


64n2
42
= y2  16
24. (5g  2h)(5g + 2h) = (5g)2  (2h)2
= 25g2  4h2
25. 16
()
1
3
= 99 —89
42
 16
= 64 
20. —2  c —2 + c = —2

1
3
1 2
—3
1
—9
⋅ 24 = (20  4)(20 + 4)
= 202  42
33. a. (x + 50)2 = x 2 + 2(x)(50) + 502
= x2 + 100x + 2500
A polynomial that represents the area of the house after
the renovation is ( x2 + 100x + 2500 ) square feet.
b. x 2 + 100x + 2500 = (15)2 + 100(15) + 2500
= 225 + 1500 + 2500
= 4225
The area of the house after the renovation is
4225 square feet.
The original area of the house was 50 2 = 2500 square feet.
So, the area of the renovation is
4225  2500 = 1725 square feet.
= 400  16
= 384
26. 33
⋅ 27 = (30 + 3)(30  3)
= 302  32
= 900  9
= 891
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Algebra 1
Worked-Out Solutions
393
Chapter 7
34. a. (100  x)(100 + x) = 1002  x 2
37. x 2  121 ts the product side of the sum and difference
= 10,000  x2
The product that represents the area of the new parking lot
is ( 10,000  x2 ) square feet.
b. The area of the parking lot decreases. The original area is
10,000 square feet. So, the new area is the original area
decreased by x2.
c. 10,000  x 2 = 10,000  212
= 10,000  441
= 9559
When x = 21, the area of the new parking lot is
9559 square feet.
35. a. The Punnett square shows four possible gene
combinations of the offspring. Of these combinations, one
results in albino coloring. So, —14 = 25% of the possible
gene combinations result in albino coloring.
b. (0.5N +
0.5a)2
=
(0.5N )2
+ 2(0.5N )(0.5a) +
=
0.25N 2
+ 0.5Na + 0.25a2
(0.5a)2
The coefcients show that 25% + 50% = 75% of the
possible gene combinations result in normal coloring,
while 25% of the possible gene combinations result in
albino coloring.
36. a. A = π r2
pattern, so working backward, a and b are the square roots of
a2 and b2.
—
—
√ x2 = x and √ 121 = 11
So, x2  121 = (x + 11)(x  11).
38. The Punnett square shows four possible gene combinations
of the offspring. Of these combinations, three result in
green pods. So, —34 = 75% of the possible gene combinations
result in green pods.
Alternately, the coefcients of the polynomial model show
that GG accounts for 25% of the possible gene combinations,
and Gy accounts for 50% of the possible gene combinations.
So, 25% + 50% = 75% of the possible gene combinations
result in green pods.
39. ( x 2 + 1 )( x 2  1 ) = ( x 2 )2 = 12
= x4  1
40. ( y3 + 4 )2 = ( y3 )2 + 2( y3 )(4) + 42
= y6 + 8y3 + 16
41. ( 2m2  5n2 )2 = ( 2m2 )2  2( 2m2 )( 5n2 ) + ( 5n2 )2
= 22m2 ⋅ 2  20m2n2 + 52n2 ⋅ 2
= 4m4  20m2n2 + 25n4
42. ( r 3  6t 4 )( r 3 + 6t 4 ) = ( r 3 )2  ( 6t 4 )2
= π (6  x)2
= r 3 ⋅ 2  62t 4 ⋅ 2
= π ( 62  2(6)(x) + x2 )
= r 6  36t 8
= π ( 36  12x + x2 )
= π (36)  π (12x) + π ( x2 )
= 36π  12π x +
π x2
= π x2  12π x + 36π
A polynomial that represents the area of your pupil is
( π x2  12π x + 36π ) square millimeters.
b. Let x = 4. Find the area of your pupil before entering
the room.
π x2  12π x + 36π = π (4)2  12π (4) + 36π
= π (16)  48π + 36π
= 16π  48π + 36π
= 4π
The area of your pupil before entering the room is
4π millimeters.
Let x = 2. Find the area of your pupil after entering
the room.
( )
1
43. Your friend is incorrect. The expression 4 —3
(
as 4 +
1
—3
not
can be written
) . However, using the square of a binomial pattern
( ) ( ) = 16 + — + —, which is
2
results in 42 + 2(4) —13 + —13
18 —79 ,
2
16 —19 .
2
8
3
1
9
44. Sample answer: One way to modify the dimensions of the
lake is by increasing one dimension by 4 and decreasing
the other dimension by 4, which can be modeled by
(x + 4)(x  4). This follows a sum and difference pattern.
A second way to modify the dimensions of the lake is by
increasing both dimensions by 2, which can be modeled by
(x + 2)2. This follows a square of a binomial pattern. A third
way to modify the dimensions of the lake is by decreasing
both dimensions by 2, which can be modeled by (x  2)2.
This also follows a square of a binomial pattern, but this one
has subtraction.
π x2  12π x + 36π = π (2)2  12π (2) + 36π
= π (4)  24π + 36π
= 4π  24π + 36π
= 16π
The area of your pupil after entering the room is 16π
16π
millimeters. So, the area of your pupil is — = 4 times
4π
greater after entering the room than before.
394
Algebra 1
Worked-Out Solutions
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Chapter 7
45. Let 9x 2  48x + k = a2  2ab + b 2.
So,
50. 49s + 35t = 7(7s + 5t)
9x2 = a2
—
51. 15x  10y = 5(3x  2y)
—
√9x2 = √a2
3x = a.
Then,
Finally,
7.4 Explorations (p. 377)
48x = 2ab
48x = 2(3x)b
48x = 6xb
48x 6xb
—=—
6x
6x
8 = b.
(x  1)(x  3) = 0 ← Factored
Form
1. a. C; 5;
F
O
I
L
x(x) + x(3) + (1)(x) + (1)(3) = 0
x2  3x  x + 3 = 0
x2  4x + 3 = 0 ← Standard
Form, C
k = b2
k = 82
k = 64.
3
So, when k = 64, 9x2  48x + k = 9x2  48x + 64 is the
square of the binomial 3x  8.
46. (x + 1)3 = (x + 1)(x + 1)(x + 1)
= [x(x + 1) + 1(x + 1)](x + 1)
= [x(x) + x(1) + 1(x) + 1(1)](x + 1)
= ( x2 + x + x + 1 )(x + 1)
= ( x2 + 2x + 1 )(x + 1)
= x2(x + 1) + 2x(x + 1) + 1(x + 1)
= x2(x) + x2(1) + 2x(x) + 2x(1) + 1(x) + 1(1)
= x3 + ( x2 + 2x2 ) + (2x + x) + 1
= x3 + 3x2 + 3x + 1
(x + 2)3 = (x + 2)(x + 2)(x + 2)
= [x(x + 2) + 2(x + 2)](x + 2)
= [x(x) + x(2) + 2(x) + 2(2)](x + 2)
= ( x2 + 2x + 2x + 4 )(x + 2)
= ( x2 + 4x + 4 )(x + 2)
= x2(x + 2) + 4x(x + 2) + 4(x + 2)
= x2(x) + x2(2) + 4x(x) + 4x(2) + 4(x) + 4(2)
= x3 + 2x2 + 4x2 + 8x + 4x + 8
= x3 + 6x2 + 12x + 8
So, (a + b)3 = a3 + 3a2b + 3ab2 + b3.
47. Sample answer: The statement (a + b)(a  b) < (a  b)2 <
(a + b)2 is true when a = 3 and b = 4.
Check (a + b)(a  b) < (a  b)2
?
(3 + 4)(3  4) < (3  4)2
?
(7)(1) < (1)2
7 <
1
< (a + b)2
?
< (3 + 4)2
?
<
(7)2
<
3
x2  4x = 3 ← Nonstandard
Form, 5
So, the equation (x  1)(x  3) = 0 is equivalent to
x2  4x + 3 = 0 (C) and x2  4x = 3 (5).
(x  2)(x  3) = 0 ← Factored
Form
b. D; 1;
F
O
I
L
x(x) + x(3) + (2)(x) + (2)(3) = 0
x2  3x  2x + 6 = 0
x2  5x + 6 = 0 ← Standard
Form, D
6
6
x2  5x = 6 ← Nonstandard
Form, 1
So, the equation (x  2)(x  3) = 0 is equivalent to
x2  5x + 6 = 0 (D) and x2  5x = 6 (1).
(x + 1)(x  2) = 0 ← Factored Form
c. A; 3;
F
O
I
L
x(x) + x(2) + 1(x) + 1(2) = 0
x2  2x + x  2 = 0
x2  x  2 = 0 ← Standard Form, A
+2
+2
x2  x = 2 ← Nonstandard
Form, 3
So, the equation (x + 1)(x  2) = 0 is equivalent to
x2  x  2 = 0 (A) and x2  x = 2 (3).
49 ✓
Maintaining Mathematical Prociency
48. 12y  18 = 6(2y  3)
49. 9r + 27 = 9(r + 3)
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Algebra 1
Worked-Out Solutions
395
Chapter 7
(x  1)(x + 2) = 0 ← Factored Form
d. B; 4;
F
O
I
L
x(x) + x(2) + (1)(x) + (1)(2) = 0
d.
x2 + 2x  x  2 = 0
x2 + x  2 = 0 ← Standard
Form, B
x(x + 1) = 2 ← Nonstandard
Form, 4
x(x) + x(1) = 2
x2
+x=2
e.
x2 + x  2 = 2  2
So, the equation (x  1)(x + 2) = 0 is equivalent to
x2 + x  2 = 0 (B) and x(x + 1) = 2 (4).
(x + 1)(x  3) = 0 ← Factored Form
F
O
I
L
x(x) + x(3) + 1(x) + 1(3) = 0
x2  2x  3 = 0 ← Standard Form, E
(x  1)2 = 4 ← Nonstandard
Form, 2
x2  2(x)(1) + 12 = 4
(x − 4)(x − 5) = 0
(3)(4) ≠ 0
False
(2)(3) ≠ 0
False
3
4
5
6
(1)(2) ≠ 0
False
(0)(1) = 0
True
(1)(0) = 0
True
(2)(1) ≠ 0
False
x=
1
2
(x − 5)(x − 6) = 0
(4)(5) ≠ 0
False
(3)(4) ≠ 0
False
3
4
5
6
f.
x=
1
2
(x − 6)(x − 1) = 0
(5)(0) = 0
True
(4)(1) ≠ 0
False
3
4
5
6
(0)(5) = 0
True
If the product of two values is 0, then at least one of the
values must be 0. If (x  a) is a factor of an equation, then
x = a is a solution.
3. a. When you add 0 to a number n, you get n. Adding 0 does
x2  2x + 1 = 4
not change the value, but adding 1 increases the value.
4
x2  2x  3 = 0 ← Standard Form, E
So, the equation (x + 1)(x  3) = 0 is equivalent to
x2  2x = 3 (E) and (x  1)2 = 4 (2).
b. If the product of two numbers is 0, then at least one of the
numbers is 0. The product of 0 and any number is always
0, never 1.
c. The square of 0 is equal to itself, and the square of 1 is
equal to itself. Both are true: 02 = 0 and 12 = 1.
2.
a.
b.
c.
396
2
(3)(2) ≠ 0 (2)(3) ≠ 0 (1)(4) ≠ 0
False
False
False
x2  3x + x  3 = 0
4
1
(2)(3) ≠ 0 (1)(2) ≠ 0 (0)(1) = 0 (1)(0) = 0
False
False
True
True
x2 + x  2 = 0 ← Standard
Form, B
e. E; 2;
x=
x=
1
2
(x − 1)(x − 2) = 0
(0)(1) = 0
True
(1)(0) = 0
True
d. When you multiply a number n by 1, you get n. The
3
4
5
6
(2)(1) ≠ 0
False
(3)(2) ≠ 0
False
(4)(3) ≠ 0
False
(5)(4) ≠ 0
False
x=
1
2
(x − 2)(x − 3) = 0
(1)(2) ≠ 0
False
(0)(1) = 0
True
3
4
5
6
(1)(0) = 0
True
(2)(1) ≠ 0
False
(3)(2) ≠ 0
False
(4)(3) ≠ 0
False
x=
1
2
(x − 3)(x − 4) = 0
(2)(3) ≠ 0
False
(1)(2) ≠ 0
False
3
4
5
6
(0)(1) = 0
True
(1)(0) = 0
True
(2)(1) ≠ 0
False
(3)(2) ≠ 0
False
Algebra 1
Worked-Out Solutions
product of any number and 1 is the number. On the other
hand, the product of any number and 0 is 0.
e. When you multiply a number n by 0, you get 0. The
product of any number and 0 is 0. On the other hand, the
product of any number and 1 is the number.
f. The opposite of 0 is equal to itself. Zero is neither positive
nor negative. So, it is its own opposite. The opposite of 1,
however, is 1.
4. When an equation is in factored form, the product of the
factors equals 0. So, set each polynomial factor equal to 0,
and solve.
5. b; If the product of two numbers is 0, then at least one of
the numbers is 0. This property is called the Zero-Product
Property because it describes what happens when you have
a product that is equal to zero. The Zero-Product Property
is used in algebra to solve polynomial equations by setting
each polynomial factor equal to 0. It is important because it
provides an easy way to solve polynomial equations.
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Chapter 7
7.4 Monitoring Progress (pp. 378–380)
4. (3s + 5)(5s + 8) = 0
1. x(x  1) = 0
x=0
or
3s + 5 =
x1= 0
5
+1 +1
x(x  1) = 0
?
1(1  1) = 0
?
1(0) = 0
0=0✓
or
t+2=
2
0
2
(
0=0✓
z6= 0
+4 +4
+6 +6
z= 4
z= 6
0=0✓
The roots are z = 4 and z = 6.
)
()
8
5
The roots are s = — and s = —.
3
5
3. (z  4)(z  6) = 0
Check (z  4)(z  6) = 0
?
(4  4)(4  6) = 0
?
0(2) = 0
)
( )
?
8
8
3 — + 5 5 — + 8 = 0
5
5
?
24
— + 5 (8 + 8) = 0
5
?
1
— (0) = 0
5
0=0✓
The roots are t = 0 and t = 2.
or
(
[ ( ) ][ ( ) ]
3t(t + 2) = 0
?
3(2)(2 + 2) = 0
?
6(0) = 0
0=0✓
z4= 0
8
(3s + 5)(5s + 8) = 0
t=2
Check 3t(t  2) = 0
?
3(0)(0 + 2) = 0
?
0(2) = 0
8
?
5
5 — + 8 = 0
3
?
25
(5 + 5) — + 8 = 0
3
1 ?
0 — = 0
3
0=0✓
5
3 — + 5
3
2. 3t(t + 2) = 0
3t = 0
5
0
[ ( ) ][ ( ) ]
0=0✓
The roots are x = 0 and x = 1.
3t 0
—=—
3
3
t=0
5s + 8 =
or
3s =  5
5s =  8
5s 8
3s 5
—=—
—=—
3
5
5
3
8
5
s = —
s = —
3
5
Check
(3s + 5)(5s + 8) = 0
x= 1
Check x(x  1) = 0
?
0(0  1) = 0
?
0(1) = 0
0
(b + 7)2 = 0
5.
(b + 7)(b + 7) = 0
(z  4)(z  6) = 0
?
(6  4)(6  6) = 0
?
2(0) = 0
0=0✓
b+7=
7
0
b+7=
or
7
7
b = 7
0
7
b = 7
Check (b + 7)2 = 0
?
(7 + 7)2 = 0
?
(0)2 = 0
0=0✓
The equation has repeated roots of b = 7.
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Algebra 1
Worked-Out Solutions
397
Chapter 7
6. (d  2)(d + 6)(d + 8) = 0
d2= 0
d+6=
or
+2 +2
6
d=2
0 or
6
d+8=
8
d = 6
0
8
d = 8
Check (d  2)(d + 6)(d + 8) = 0
?
(2  2)(2 + 6)(2 + 8) = 0
?
0(8)(10) = 0
0=0✓
7. The GCF of 8 and 24 is 8. The GCF of y2 and y is y. So,
the greatest common monomial factor of the terms is 8y.
So, 8y2  24y = 8y( y  3).
2x = 1
4x2 = 2x
1 2 ?
1
4 — =2 —
2
2
1 ?
4 — =1
0=0✓
4
1=1✓
1
The roots are x = 0 and x = —.
2
0
()
1
y = —2 (x + 4)(x  4)
⋅ ( — )(x + 4)(x  4)
1
2
0 = (x + 4)(x  4)
5
x+4=
a = 5
Check a2 + 5a = 0
?
(0)2 + 5(0) = 0
?
0+0=0
()
()
11. Let y = 0.
2(0) = 2
5
+1
1
+ 5a = 0
a2 + 5a = 0
?
(5)2 + 5(5) = 0
?
25  25 = 0
0=0✓
0=0✓
The roots are a = 0 and a = 5.
4
0
or
x4= 0
4
+4 +4
x = 4
x= 4
So, the width of the entrance at ground level is
∣  4  4 ∣ = 8 feet.
7.4 Exercises (pp. 381–382)
Vocabulary and Core Concept Check
9. 3s2  9s = 0
1. The product of 3x and x  6 equals 0. According to the
3s(s  3) = 0
or
+1
0
0 = —2 (x + 4)(x  4)
a(a + 5) = 0
a+5=
2x  1 =
Check 4x2 = 2x
?
4(0)2 = 2(0)
?
4(0) = 0
The roots are d = 2, d = 6, and d = 8.
3s = 0
3s 0
—=—
3
3
s=0
or
1
2
1
x=—
2
(d  2)(d + 6)(d + 8) = 0
?
(8  2)(8 + 6)(8 + 8) = 0
?
(10)(2)(0) = 0
or
2x(2x  1) = 0
2x
2
0=0✓
a=0
4x2  2x = 0
—=—
(d  2)(d + 6)(d + 8) = 0
?
(6  2)(6 + 6)(6 + 8) = 0
?
(8)(0)(2) = 0
8.
4x2  2x = 2x  2x
2x = 0
2x 0
—=—
2
2
x=0
0=0✓
a2
4x2 = 2x
10.
s3=
+3
Zero-Product Property, at least one of the factors equals 0.
So, let 3x = 0 and x  6 = 0. Then solve each equation to
get the solutions x = 0 and x = 6.
0
+3
s=3
Check 3s2  9s = 0
?
3(0)2  9(0) = 0
?
00=0
0=0✓
3s2  9s = 0
?
3(3)2  9(3) = 0
?
3(9)  27 = 0
?
27  27 = 0
0=0✓
The roots are s = 0 and s = 3.
398
Algebra 1
Worked-Out Solutions
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Chapter 7
2. The one that is different is “Find the value of k for which
(2k + 4) + (k  3) = 0.”
6. 2v(v + 1) = 0
2v = 0
or
2v
0
—=—
2
2
v=0
(2k + 4) + (k  3) = 0
(2k + k) + (4  3) = 0
3k + 1 =
1
0
v+1=
1
v = 1
1
7. (s  9) (s  1) = 0
(2k + 4)(k  3) = 0
8. ( y + 2) ( y  6) = 0
4
0
s9=
+9
4
2
k=3
y6=
or
+6
0
+6
y=6
9. (2a  6) (3a + 15) = 0
2a  6 = 0
x+7=
7
—=—
The roots are a = 3 and a = 5.
x = 7
10. (4q + 3) (q + 2) = 0
4q + 3 =
4. r(r  10) = 0
3
0
r = 10
q+2=
or
3
2
3
4
3
q = —
4
3
The roots are q = — and q = 2.
4
0
t=5
The roots are t = 0 and t = 5.
2
q = 2
4q
4
5. 12t(t  5) = 0
+5
0
—=—
The roots are r = 0 and r = 10.
t5=
0
4q = 3
+ 10 + 10
+5
3a 15
3
3
a = 5
2a 6
2
2
a=3
7
0
 15
3a = 15
—=—
The roots are x = 0 and x = 7.
or
 15
2a = 6
0
r  10 =
3a + 15 =
or
+6 +6
3. x(x + 7) = 0
12t = 0
0
12t
—=—
12
12
t=0
0
2
y = 2
Monitoring Progress and Modeling with Mathematics
or
0
+1
The roots are y = 2 and y = 6.
The roots are k = 2 and k = 3.
r=0
+1
s=1
y+2=
+3 +3
2k 4
—=—
2
2
k = 2
or
s1=
or
s=9
2k = 4
x=0
0
+9
The roots are s = 9 and s = 1.
k3= 0
or
1
The roots are v = 0 and v = 1.
3k = 1
3k 1
—=—
3
3
1
k = —
3
1
The solution is k = —.
3
2k + 4 =
0
(5m + 4)2 = 0
11.
(5m + 4) (5m + 4) = 0
5m + 4 =
4
0
4
5m = 4
or
5m + 4 =
4
0
4
5m = 4
5m 4
—=—
5
5
4
m = —
5
5m 4
—=—
5
5
4
m = —
5
4
The equation has repeated roots of m = —.
5
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Algebra 1
Worked-Out Solutions
399
Chapter 7
(h  8)2 = 0
12.
(r  4)(r  4)(r + 8) = 0
(h  8) (h  8) = 0
h8=
0
h8=
or
+8 +8
+8
h=8
r4=
0
+8
+4
h=8
0
7g=
or
+ 2g
+g
3 = 2g
+4
0
w=0
or w  6 =
+6
15  5c =
2
2
4d = 2
0
 15
2
4d = 2
0
or
2
0
5
p=0
+3
0
z=1
p+7= 0
+3
2p = 3
2p 3
—=—
2
2
3
p=—
2
Algebra 1
Worked-Out Solutions
0
or c + 6 =
5
6
0
6
15
5
—=—
5
5
—=—
5c
5
c
1
c = 1
)
2
3
0
or
2
6
1
c=6
7 7
p = 7
or
2
3
3 2
3
— —n = — ( 6 )
2 3
2
2
1
n
1
2
6 + —n = 0
3
6
6
—n = 6
n = 2
⋅
—=—
n=2
3
The roots are p = 0, p = —, and p = 7.
2
400
5
c = 6
2n=
+1 +1
or
5c + 5 =
5c = 5
(
16. 5p(2p  3)(p + 7) = 0
0
or
20. (2  n) 6 + —n (n  2)= 0
The roots are z = 0, z = 2, and z = 1.
2p  3 =
0
+6
w=6
c=3
z1=
z = 2
5p
5
+6
5c = 15
2
2
—=—
w6=
or
The roots are c = 3, c = 1, and c = 6.
15. z (z + 2)(z  1) = 0
z+2=
0
 15
5c
5
4d 2
—=—
4
4
1
d = —
2
1
1
The roots are d = — and d = —.
2
2
or
0
+6
—=—
4d 2
—=—
4
4
1
d=—
2
5p = 0
r = 8
19. (15  5c)(5c + 5)(c + 6) = 0
2 + 4d =
or
8
One root of the equation is w = 0. The equation also has
repeated roots of w = 6.
14. (2  4d ) (2 + 4d ) = 0
0
0
8
w=6
3
The roots are g = — and g = 7.
2
or
r+8=
w(w  6)(w  6) = 0
7=g
3
2
z=0
or
18. w(w  6)2 = 0
+g
—=g
2
0
+4
r=4
3 2g
—=—
2
2
2  4d =
r4=
or
One root of the equation is r = 8. The equation also has
repeated roots of r = 4.
13. (3  2g) (7  g) = 0
3  2g =
0
+4
r=4
The equation has repeated roots of h = 8.
+ 2g
(r  4)2(r + 8) = 0
17.
n2= 0
+2 +2
n=2
⋅
n = 9
One root of the equation is n = 9. The equation also has
repeated roots of n = 2.
21. y = (x  8)(x + 8)
0 = (x  8)(x + 8)
x8=
+8
0
+8
x=8
or
x+8=
8
0
8
x = 8
The x-coordinates of the points where the graph crosses the
x-axis are the roots x = 8 and x = 8.
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Chapter 7
22. y = ( x + 1 )( x + 7 )
32. 6m2 + 12m = 0
0 = ( x + 1 )( x + 7 )
6m( m + 2 ) = 0
x+1=0
1
x+7=0
or
1
7
x = 1
6m = 0
6m 0
—=—
6
6
m=0
7
x = 7
The x-coordinates of the points where the graph crosses the
x-axis are the roots x = 1 and x = 7.
+14
5c = 0
5c 0
—=—
5
5
c=0
+5
x = 14
+5
x=5
The x-coordinates of the points where the graph crosses the
x-axis are the roots x = 14 and x = 5.
24.
y = 0.2( x + 22 )( x  15 )
0 = 0.2( x + 22 )( x  15 )
0.2( x + 22 )( x  15 )
0
— = ——
0.2
0.2
0 = ( x + 22 )( x  15 )
x + 22 = 0
22
or
+15
x = 22
2q 0
2
2
q=0
9=q
3n2
 9n = 9n  9n
3n2
 9n = 0
3n( n  3 ) = 0
3n = 0
or
n3=0
3n 0
—=—
3
3
n=0
+3 +3
n=3
The roots are n = 0 and n = 3.
36.
28r = 4r2
28r + 28r = 4r2 + 28r
0 = 4r2 + 28r
0 = 4r( r + 7 )
31. 4p2  p = 0
p( 4p  1 ) = 0
or
+q +q
3n2 = 9n
35.
28. 20x3 + 30x2 = 10x2( 2x + 3 )
p=0
9q=0
The roots are q = 0 and q = 9.
27. 3y3  9y2 = 3y2( y  3 )
30. 12a4 + 8a = 4a( 3a3 + 2 )
or
—=—
26. 6d2  21d = 3d( 2d  7 )
29. 5n6 + 2n5 = n5( 5n + 2 )
5
2q( 9  q ) = 0
The x-coordinates of the points where the graph crosses the
x-axis are the roots x = 22 and x = 15.
25. 5z2 + 45z = 5z( z + 9 )
5
2q = 0
+15
x = 15
5 + 2c = 0
34. 18q  2q2 = 0
x  15 = 0
22
or
2c = 5
2c 5
—=—
2
2
5
c=—
2
5
The roots are c = 0 and c = —.
2
x5=0
+14
m = 2
5c( 5 + 2c ) = 0
0 = ( x  14 )( x  5 )
or
2 2
33. 25c + 10c2 = 0
1( 0 ) = 1 ⋅ [ ( x  14 )( x  5 ) ]
x  14 = 0
m+2=0
The roots are m = 0 and m = 2.
y = ( x  14 )( x  5 )
0 = ( x  14 )( x  5 )
23.
or
4p  1 = 0
+1 +1
4p = 1
4p 1
—=—
4
4
1
p=—
4
4r = 0
4r 0
—=—
4
4
r=0
or
r+7=0
7
7
r = 7
The roots are r = 0 and r = 7.
1
The roots are p = 0 and p = —.
4
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Algebra 1
Worked-Out Solutions
401
Chapter 7
37. The other factor, 6x, should also be set equal to 0. Then
each equation should be solved to get two roots of the
original equation.
6x( x + 5 ) = 0
x=0
x+5=0
or
6x 0
—=—
6
6
x=0
5 5
x = 5
b. Let x = 0.
2
y = —( x + 315 )( x  315 )
315
2
y = —( 0 + 315 )( 0  315 )
315
2
y = —( 315 )( 315 )
315
= ( 2 )( 315 )
= 630
The roots are x = 0 and x = 5.
So, the height of the arch is 630 feet.
38. Each side of the equation cannot be divided by y because
y could be 0, and division by 0 is undened. Instead, 21y
should be subtracted from each side, so that the equation
is 3y2  21y = 0, and the greatest monomial factor should
be factored out on the left side of the equal sign. Then each
factor should be set equal to zero, and each equation should
be solved to nd both roots of the original equation.
41. y = 16x2 + 4.8x
0 = 16x2 + 4.8x
0 = x( 16x + 4.8 )
x=0
or
4.8
3y2  21y = 21y  21y
3y2  21y = 0
3y( y  7 ) = 0
Because y = 0, the roots are the x-values that represent the
times when the penguin is at water level. So, when x = 0, the
penguin emerges from the water. Then it is above the water
until 0.3 second later, the moment it breaks the surface of the
water again after the leap. So, the penguin is airborne for
0.3 second with each leap.
y7=0
or
3y 0
3
3
y=0
—=—
+7 +7
y=7
The roots are y = 0 and y = 7.
42. The point of intersection on the positive x-axis is farther
39. Let y = 0.
from the origin than the point of intersection on the negative
x-axis. So, the positive solution to the equation is 5, because
it is the one with the greater absolute value. Therefore, the
equation is y = ( x  5 )( x + 3 ).
11
y = —( x  4 )( x  24 )
50
11
0 = —( x  4 )( x  24 )
50
50( )
50 11 (
— 0 = — — x  4 )( x  24 )
11
11 50
0 = ( x  4 )( x  24 )
( )
x4=0
+4
43. The graph has two x-intercepts because x-intercepts occur
when y = 0, and this equation has two roots (one of which
is a repeated root) when y = 0.
x  24 = 0
or
+4
+24
x=4
44. no; The equation y = ( x  a )( x  b ) will not have two
+24
x-intercepts if a = b. In this case, there is a repeated root,
and the graph will only have one x-intercept.
x = 24
So, the width of the tunnel at ground level is ∣ 24  4 ∣ = 20 feet.
40. a. Let y = 0.
2
y = —( x + 315 )( x  315 )
315
2
0 = —( x + 315 )( x  315 )
315
315
315
2
—( 0 ) = — — ( x + 315 )( x  315 )
2
2
315
0 = ( x + 315 )( x  315 )
(
x + 315 = 0
315
)
or
x  315 = 0
315
+315 +315
x = 315
x = 315
So, the width of the arch at ground level is
∣ 315  315 ∣ = 630 feet.
402
4.8
16x = 4.8
16x
4.8
— = —
16
16
x = 0.3
3y2 = 21y
3y = 0
16x + 4.8 = 0
Algebra 1
Worked-Out Solutions
45. no; The equation ( x2 + 3 )( x4 + 1 ) = 0 does not have any
real roots. Roots will occur if x3 + 3 = 0 or x4 + 1 = 0.
However, solving these equations results in x2 = 3 or
x4 = 1, and even powers of any number cannot be
negative.
x3 + 3 = 0
x4 + 1= 0
3
3
1
1
x3
= 3
x4
= 1
46. Sample answer: A polynomial equation of degree 4 that only
has three roots must have a repeated root. So, one possible
equation is ( x  1 )( x  2 )2( x  3 ) = 0.
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Chapter 7
52. 48
47. a. ( x + y )( 2x  y ) = 0
x+y=0
x+yy=0y
1( 48 ) = 48
2( 24 ) = 48
2x  y = 0
or
2x  y + y = 0 + y
x = y
2x = y
2x
2
y
2
1
x = —y
2
1
So, the roots of the equation occur when x = y and x = —y.
2
—=—
b. ( x2  y2 )( 4x + 16y ) = 0
x2  y2 = 0
4x + 16y = 0
or
x2  y2 + y2 = 0 + y2
4x  4x + 16y = 0  4x
x2 = y2
—
16y = 4x
16y 4x
4
4
4y = x
—
± √x2 = ± √ y2
—= —
x=±y
So, the roots of the equation occur when x = ± y and x = 4y.
48.
( 4x5
 16 )
( 3x
 81 ) = 0
4x5  16 = 0
or
+16 +16
3x  81 = 0
+81
+81
4x5 = 16
3x = 81
4x5 = 42
3x = 34
x5=2
x=4
+5 +5
x=7
The solutions are x = 7 and x = 4.
3( 16 ) = 48
4( 12 ) = 48
6( 8 ) = 48
So, the factor pairs of 48 are 1 and 48, 2 and 24, 3 and 16, 4
and 12, and 6 and 8.
7.1–7.4 What Did You Learn? (p. 383)
1. Sample answer: Use x + 4 to represent each side of the
square. So, an expression that represents the area is
( x + 4 )2. Then use the square of a binomial pattern to
simplify this expression into a polynomial. As another method,
you could calculate the area of each shaded region and add
those areas together.
2. Sample answer: The solutions are the constant terms with
the opposite sign. This method does not work when the
coefcient is not 1.
7.1–7.4 Quiz (p. 384)
1. The polynomial 8q3 is in standard form.
The only term has a degree of 3. So, the degree of the
polynomial is 3.
The leading coefcient is 8.
The polynomial has 1 term. So, it is a monomial.
2. You can write the polynomial 9 + d 2  3d in standard form
as d 2  3d + 9.
The greatest degree is 2. So, the degree of the polynomial is 2.
Maintaining Mathematical Prociency
49. 10
1( 10 ) = 10
2( 5 ) = 10
So, the factor pairs of 10 are 1 and 10, and 2 and 5.
50. 18
1( 18 ) = 18
2( 9 ) = 18
3( 6 ) = 18
So, the factor pairs of 18 are 1 and 18, 2 and 9, and 3 and 16.
51. 30
1( 30 ) = 30
2( 15 ) = 30
3( 10 ) = 30
The leading coefcient is 1.
The polynomial has 3 terms. So, it is a trinomial.
2
3
5 m6 + 2 m4.
as  —
—
6
3
The greatest degree is 6. So, the degree of the polynomial is 6.
5
The leading coefcient is —.
6
The polynomial has 2 terms. So, it is a binomial.
4. You can write the polynomial 1.3z + 3z4 + 7.4z2 in
standard form as 3z4 + 7.4z2  1.3z.
The greatest degree is 4. So, the degree of the polynomial is 4.
The leading coefcient is 3.
The polynomial has 3 terms. So, it is a trinomial.
5. ( 2x2 + 5 ) + ( x2 + 4 ) = 2x2  x2 + 5 + 4
= ( 2x2  x2 ) + ( 5 + 4 )
5( 6 ) = 30
So, the factor pairs of 30 are 1 and 30, 2 and 15, 3 and 10,
and 5 and 6.
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5
6
3. You can write the polynomial — m4  — m6 in standard form
= x2 + 9
The sum is
x2
+ 9.
Algebra 1
Worked-Out Solutions
403
Chapter 7
6. ( 3n2 + n )  ( 2c2  7 ) = 3n2 + n  2n2 + 7
The difference is
5n2
5x(x  3) = 0
= 5n2 + n + 7
5x = 0
5x 0
—=—
5
5
x=0
+ n + 7.
)(
(
= p2 + 4p  p2 + 3p  15
p2 + 4p
7.
p2  3p + 15
)
3=0
8g=0
The difference is 2p2 + 7p  15.
8. ( a2  3ab + b2 ) + ( a2 + ab + b2 )
= a2  a2  3ab + ab + b2 + b2
= ( a2  a2 ) + ( 3ab + ab ) + ( b2 + b2 )
= 0  2ab + 2b2
+g +g
+g +g
8=g
8=g
The equation has repeated roots of g = 8.
17. (3p + 7)(3p  7)(p + 8) = 0
7
The sum is 2ab + 2b2.
9. ( w + 6 )( w + 7 ) = w( w + 7 ) + 6( w + 7 )
= w( w ) + w( 7 ) + 6( w ) + 6( 7 )
= w2 + 7w + 6w + 42
= w2 + 13w + 42
+ 13w + 42.
8g=0
or
3p + 7 =
= 2ab + 2b2
0
or
3p  7 =
7
+7
0
or
+7
O
I
L
= 3( 2d ) + 3( 5 ) + ( 4d )( 2d ) + ( 4d )( 5 )
= 6d  15  8d 2 + 20d
= 8d 2 + ( 6d + 20d )  15
8
3p = 7
7
3p
— = —
3
3
7
p = —
3
3p = 7
3p 7
—=—
3
3
7
p=—
3
7
7
The roots are p = —, p = —, and p = 8.
3
3
3y = 0 or y  8 = 0
or
0
3y
—=—
+8
+8
3
3
y=0
y=8
The product is 8d 2 + 26d  15.
y2 + 2y  3
y+9
×
——
2 + 18y  27
9y
y3 + 2y2  3y
——
3 + 11y2 + 15y  27
y——
The product is y3 + 11y2 + 15y  27.
12. (3z  5)(3z + 5) = (3z)2  52
= 9z2  25
The product is 9z2  25.
13. (t + 5)2 = t 2 + 2(t)(5) + 52
=t 2
+ 10t + 25
The product is t 2 + 10t + 25.
14. (2q  6)2 = (2q)2  2(2q)(6) + 62
= 4q2  24q + 36
The product is 4q2  24q + 36.
404
0
8
p = 8
2y + 1 = 0
1
= 8d 2 + 26d  15
11.
p+8=
18. 3y(y  8)(2y + 1) = 0
10. ( 3  4d )( 2d  5 )
F
+3 +3
16. (8  g)(8  g) = 0
= 2p2 + 7p  15
The product is
x3=0
or
The roots are x = 0 and x = 3.
= ( p2  p2 ) + ( 4p + 3p )  15
w2
15. 5x2  15x = 0
= ( 3n2  2n2 ) + n + 7
Algebra 1
Worked-Out Solutions
1
The roots are y = 0, y = 8, and y = —.
2
1
2y = 1
2y 1
—=—
2
2
1
y = —
2
19. a. P = 2ℓ + 2w
= 2(x + 72 + x) + 2(x + 48 + x)
= 2(2x + 72) + 2(2x + 48)
= 2(2x) + 2(72) + 2(2x) + 2(48)
= 4x + 144 + 4x + 96
= 8x + 240
A polynomial that represents the perimeter of the blanket
including the fringe is (8x + 240) inches.
b. A = ℓ w
= (x + 72 + x)(x + 48 + x)
= (2x + 72)(2x + 48)
F
O
I
L
= 2x(2x) + 2x(48) + 72(2x) + 72(48)
= 4x2 + 96x + 144x + 3456
= 4x2 + 240x + 3456
A polynomial that represents the area of the blanket
including the fringe is (4x2 + 240x + 3456) square inches.
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Chapter 7
c. P = 8x + 240
Section 7.5
= 8(4) + 240
7.5 Explorations (p. 385)
= 32 + 240
1. a.
= 272
A = 4x2 + 240x + 3456
= 4(4)2 + 240(4) + 3456
= 4(16) + 960 + 3456
Area = x 2  3x + 2 = (x  1)(x  2)
= 64 + 960 + 3456
b.
= 4480
When the width of the fringe is 4 inches, the perimeter of
the blanket is 272 inches and the area is 4480 square inches.
20. a. 1000(1 + r)2 = 1000(1 + r)(1 + r)
F
O
I
Area = x 2 + 5x + 4 = (x + 1)(x + 4)
L
= 1000[ 1(1) + 1(r) + r(1) + r(r) ]
c.
= 1000(1 + r + r + r2)
= 1000(r2 + 2r + 1)
= 1000(r2) + 1000(2r) + 1000(1)
= 1000r2 + 2000r + 1000
The polynomial in standard form that represents the
balance of your bank account after 2 years is
(1000r2 + 2000r + 1000) dollars.
b.
1000r2 + 2000r + 1000 = 1000(0.03)2 + 2000(0.03) + 1000
= 1000(0.0009) + 60 + 1000
Area = x 2  7x + 12 = (x  4)(x  3)
d.
= 0.9 + 60 + 1000
= 1060.9
When the interest rate is 3%, the balance after 2 years is
$1060.90.
c. 1000(1 + r3) = 1000(1 + 0.03)3
= 1000(1.03)3
Area = x 2 + 7x + 12 = (x + 4)(x + 3)
= 1000(1.092727)
2. Arrange algebra tiles that model the trinomial into a
= 1092.73
The balance after 3 years is only $1092.73. So, you do not
have enough money for the $1100 guitar.
5
y = —
216 (x  72)(x + 72)
21.
0=
216
5
—
216 (x
216
(
3. Find two integer factors of c that have a sum of b, then write
the binomial factors by adding each integer factor to x.
 72)(x + 72)
5
)
7.5 Monitoring Progress (pp. 386–388)
—
—
—
50 (0) =  5  216 (x  72)(x + 72)
1. x 2 + 7x + 6
0 = (x  72)(x + 72)
x  72 =
+ 72
0
+ 72
or
x + 72 =
 72
x = 72
0
 72
x = 72
So, the width of the bunker at ground level is
∣ 72  72 ∣ = 144 inches.
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All rights reserved.
rectangular array, use additional algebra tiles to model the
dimensions of the rectangle, then write the polynomial in
factored form using the dimensions of the rectangle.
Factors of 6
Sum of factors
1, 6
7
2, 3
5
So, x 2 + 7x + 6 = (x + 1)(x + 6).
Algebra 1
Worked-Out Solutions
405
Chapter 7
2. x 2 + 9x + 8
9.
200 = (s  30)(s  40)
200 = s(s) + s(40) + (30)(s) + (30)(40)
Factors of 8
Sum of factors
1, 8
9
200 = s 2  40s  30s + 1200
2, 4
6
200 = s 2  70s + 1200
 200
So, x 2 + 9x + 8 = (x + 1)(x + 8).
3. w 2  4w + 3
0 = (s  50)(s  20)
Factors of 3
1, 3
s  50 =
4
+ 50
Sum of factors
1, 35
5, 7
36
12
Sum of factors
So,
1, 24 2, 12 3, 8
Sum of factors
25
14
4, 6
11
10
So, x 2  14x + 24 = (x  2)(x  12).
Factors of −15
1, 15
1, 15
3, 5
3, 5
Sum of factors
14
14
2
2
+ 2x  15 = (x  3)(x + 5).
1. The signs tell you that the factors have constant terms
that are opposite signs. When factoring
x 2 + bx + c = (x + p)(x + q), if c is negative, p and q
must have opposite signs.
(x + p)(x + q), where p and q are positive is x 2 + 5x + 6. It
can be factored as (x + 2)(x + 3), where p = 2 and q = 3.
Monitoring Progress and Modeling with Mathematics
3. x 2 + 8x + 7
Factors of 7
7. y 2 + 13y  30
Factors
of −30
1, 1, 2, 2, 3, 3, 5,
30 30 15 15 10 10 6
Sum of
factors
29
29
13
13
7
7
1
5,
6
1
Factors
of −42
1, 1, 2, 2, 3, 3, 6,
42 42 21 21 14 14 7
Sum of
factors
41
19
19
 v  42 = (v + 6)(v  7).
11
11
1
6,
7
1
1, 7
8
So, x 2 + 8x + 7 = (x + 1)(x + 7).
4. z2 + 10z + 21
Sum of factors
8. v 2  v  42
41
Sum of factors
Factors of 21
So, y 2 + 13y  30 = (y  2)(y + 15).
So,
+ 20
s = 20
2. Sample answer: A trinomial that can be factored as
6. x 2 + 2x  15
v2
+ 20
0
Vocabulary and Core Concept Check
Factors of 24
So,
+ 50
s  20 =
7.5 Exercises (pp. 389–390)
 12n + 35 = (n  5)(n  7).
5. x 2  14x + 24
x2
or
The diagram shows that the side length is more than
40 meters, so a side length of 20 meters does not make sense
in this situation. The side length is 50 meters. So, the area of
the square plot of land is 50(50) = 2500 square meters.
4. n2  12n + 35
Factors of 35
0
s = 50
So, w 2  4w + 3 = (w  1)(w  3).
n2
 200
0 = s 2  70s + 1000
1, 21
3, 7
22
10
So, z2 + 10z + 21 = (z + 3)(z+ 7).
5. n2 + 9n + 20
Factors of 20
Sum of factors
1, 20
2, 10
4, 5
21
12
9
So, n2 + 9n + 20 = (n + 4)(n + 5).
6. s 2 + 11s + 30
Factors of 30
Sum of factors
1, 30
2, 15
3, 10
5, 6
31
17
13
11
So, s 2 + 11s + 30 = (s + 5)(s + 6).
406
Algebra 1
Worked-Out Solutions
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Chapter 7
7. h 2 + 11h + 18
Factors of 18
Sum of factors
16. z2 + 7z  18
1, 18
2, 9
3, 6
19
11
18
Factors
of −18
Sum of
factors
So, h 2 + 11h + 18 = (h + 2)(h + 9).
8. y 2 + 13y + 40
Factors of 40
Sum of factors
1, 40
2, 20
4, 10
5, 8
41
22
14
13
Sum of
factors
1, 4 2, 2
5
1, 22
2, 11
23
13
Sum of factors
1, 6
2, 3
5,
8
Sum of
factors
3
39
39
18
18
6
6
3
7
5
Factors 1, 1, 2, 2, 3, 3, 4, 4, 6, 6,
of −48 48 48 24 24 16 16 12 12 8 8
Sum of
47 47 22 22 13 13 8
factors
So, d 2  5d + 6 = (d  2)(d  3).
Factors of 24
Sum of factors
1, 24 2, 12 3, 8
25
14
4, 6
11
10
So, k 2  10k + 24 = (k  4)(k  6).
 17w + 72
1,
72
2,
36
3,
24
4,
18
6,
12
8,
9
Sum of
factors
73
38
27
22
18
17
So, w 2  17w + 72 = (w  8)(w  9).
14. j 2  13j + 42
Sum of factors
2
2
20. h 2 + 6h  27
Factors of −27
1, 27
1, 27
3, 9
3, 9
Sum of factors
26
26
6
6
So, h 2 + 6h  27 = (h  3)(h + 9).
Factors
of 72
Factors of 42
8
So, y 2 + 2y  48 = ( y  6)( y + 8).
12. k 2  10k + 24
21. x 2  x  20
Factors
of −20
1,
20
1,
20
2,
10
2,
10
4,
5
4,
5
Sum of
factors
19
19
8
8
1
1
So, x 2  x  20 = (x + 4)(x  5).
1, 42 2, 21 3, 14 6, 7
43
23
17
So, j 2  13j + 42 = ( j  6)( j  7).
15.
Factors 1, 1, 2, 2, 4, 4, 5,
of −40 40 40 20 20 10 10 8
19. y 2 + 2y  48
Factors of 6
x2
1
1
So, s 2 + 3s  40 = (s  5)(s + 8).
11. d 2  5d + 6
13.
4
4
18. s 2 + 3s  40
So, x 2  13x + 22 = (x  2)(x  11).
w2
11
11
So, n2 + 4n  12 = (n  2)(n + 6).
4
10. x 2  13x + 22
Sum of factors
3
3
Factors
of −12 1, 12 1, 12 2, 6 2, 6 3, 4 3, 4
So, v 2  5v + 4 = (v  1)(v  4).
Factors of 22
7
7
17. n2 + 4n  12
9. v 2  5v + 4
Sum of factors
17
17
So, z2 + 7z  18 = (z  2)(z + 9).
So, y 3 + 13y + 40 = (y + 5)(y + 8).
Factors of 4
1, 18 1, 18 2, 9 2, 9 3, 6 3, 6
+ 3x  4
Factors of −4
1, 4
1, 4
2, 2
Sum of factors
3
3
0
13
22. m2  6m  7
Factors of −7
1, 7
1, 7
Sum of factors
6
6
So, m2  6m  7 = (m + 1)(m  7).
So, x 2 + 3x  4 = (x  1)(x + 4).
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Algebra 1
Worked-Out Solutions
407
Chapter 7
23. 6t  16 + t 2 = t 2  6t  16
Factors
of −16
1, 16
Sum of
factors
So,
t2
28. Because c is 60, the factors should have opposite signs.
1, 16
2, 8
15
15
2, 8
6
6
s 2  17s  60
4, 4
Factors
of −60
0
 6t  16 = (t + 2)(t  8).
Sum of
factors
24. 7y + y 2  30 = y 2  7y  30
Factors
of −30
1, 30 1, 30 3, 10 3, 10 5, 6 5, 6
Sum of
factors
29
29
7
7
1
1
1,
60
2,
30
2,
30
3,
20
3,
20
4,
15
4,
15
5,
12
5,
12
6,
10
6,
10
59
59
28
28
17
17
11
11
7
7
4
4
So, s2  17s  60 = (s + 3)(s  20).
29.
m2 + 3m + 2 = 0
(m + 1)(m + 2) = 0
So, y 2  7y  30 = ( y + 3)( y  10).
Factors of 2
25. a. x 2  8x + 15
Factors of 15
1,
60
1, 15
3, 5
16
8
Sum of factors
1, 2
Sum of factors
3
m+1=
1
So, x 2  8x + 15 = (x  3)(x  5), and a binomial that
represents the height of the projection is (x  5) feet.
0
1
0
2
m = 2
The roots are m = 1 and m = 2.
P = 2ℓ + 2w
30.
= 2(x  3) + 2(x  5)
n2  9n + 18 = 0
= 2(8  3) + 2(8  5)
(n  6)(n  3) = 0
= 2(5) + 2(3)
Factors of 18
= 10 + 6
1, 18 2, 9 3, 6
Sum of factors
= 16
n6=
So, the perimeter of the projection is 16 feet.
+6
26. a. x 2 + x  30
Factors
of −30 1, 30 1, 30 2, 15 2, 15 5, 6 5, 6
29
2
m = 1
b. Let x = 8.
Sum of
factors
m+2=
or
29
13
13
1
1
So, + x  30 = (x  5)(x + 6), and a binomial that
represents the width of the land is (x  5) meters.
x2
b. Let x = 20.
x 2 + x  30 = 202 + 20  30
0
n3=
or
+3
n=6
0
+3
n=3
31. x 2 + 5x  14 = 0
(x  2)(x + 7) = 0
Factors of −14
1, 14
1, 14
2, 7
2, 7
Sum of factors
13
13
5
5
0
+2
or
x+7=
7
x=2
So, the area of the land is 390 square meters.
9
The roots are n = 6 and n = 3.
+2
= 390
11
+6
x2=
= 400 + 20  30
19
0
7
x = 7
The roots are x = 2 and x = 7.
27. The sum of the factors 4 and 12 is 16, not 14.
x 2 + 14x + 48
Factors of 48
Sum of factors
1, 48
2, 24
3, 16
4, 12
6, 8
49
26
19
16
14
So, x 2 + 14x + 48 = (x + 6)(x + 8).
408
Algebra 1
Worked-Out Solutions
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Chapter 7
36.
32. v 2 + 11v  26 = 0
y 2  2y  8  7 = 7  7
(v  2) (v + 13) = 0
Factors of −26
1, 26
1, 26
2, 13
2, 13
Sum of factors
25
25
11
11
v2=
0
+2
v + 13 =
or
+2
 13
v=2
y 2  2y  15 = 0
( y + 3)( y  5) = 0
Factors of −15 1, 15 1, 15 3, 5 3, 5
0
Sum of factors
 13
y+3=
v = 13
3
The roots are v = 2 and v = 13.
(t + 3)(t + 12) = 0
1, 36 2, 18 3, 12
t+3=
3
0
20
15
or t + 12 =
3
 12
t = 3
4, 9
6, 6
13
12
3
or
m4=
+4
10
10
n8=
3
+8
5
5
2
2
b 2 + 5 = 8b  10
38.
b2
+ 5  8b = 8b  8b  10
b 2  8b + 5 = 10
b 2  8b + 5 + 10 = 10 + 10
1, 15 3, 5
Sum of factors
b3=
+3
a 2 + 5a  50 = 0
0
+3
b=3
(a  5)(a + 10) = 0
or
16
8
b5=
+5
0
+5
b=5
The roots are b = 3 and b = 5.
Factors
of −50
1,
50
1,
50
2,
25
2,
25
5,
10
5,
10
Sum of
factors
49
49
23
23
5
5
a5=
0
+5
+ 11
m = 11
Factors of 15
+ 5a  20  30 = 30  30
a=5
0
+ 11
(b  3)(b  5) = 0
+8
a 2 + 5a  20 = 30
+5
m  11 =
15
b 2  8b + 15 = 0
0
n=8
or
or
+4
m=4
n = 3
a2
0
24
The roots are m = 4 and m = 11.
The roots are n = 3 and n = 8.
35.
45
Sum of factors
3, 3, 4, 4,
2,
2,
1,
Factors 1,
8 8 6
24 24 12 12
6
of −24
0
1, 44 2, 22 4, 11
Factors of 44
t = 12
(n + 3)(n  8) = 0
n+3=
+ 10  15m = 15m  15m  34
(m  4)(m  11) = 0
 12
n2  5n  24 = 0
23
y=5
m2  15m + 44 = 0
0
n2  5n  24 = 24  24
23
+5
m2  15m + 10 + 34 = 34 + 34
n2  5n = 24
Sum of
factors
0
+5
m2  15m + 10 = 34
The roots are t = 3 and t = 12.
34.
2
m2 + 10 = 15m  34
m2
37
or
3
37.
t 2 + 15t + 36 = 0
Sum of factors
y5=
2
The roots are y = 3 and y = 5.
t 2 + 15t + 36 = 36 + 36
Factors of 36
0
14
14
y = 3
t 2 + 15t = 36
33.
y 2  2y  8 = 7
a + 10 =
 10
0
 10
a = 10
The roots are a = 5 and a = 10.
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Algebra 1
Worked-Out Solutions
409
Chapter 7
39. The length of the cut picture is (x  5) inches and the width
is (x  6) inches.
b.
3
Area of
=—
13
browser window
A =ℓ w
3
24 = —
13
13
13
— (24) = —
3
3
104 = A
20 = (x  5)(x  6)
20 = x(x) + x(6) + (5)x + (5)(6)
20 =
x2
 6x  5x + 30
20 = x 2  11x + 30
1, 10
2, 5
Sum of factors
11
7
x  10 =
0 or
x1=
—
+1
x = 10
104 = 13w
104 13w
—=—
13
13
8=w
So, the screen is 6 + 7 = 13 inches by 8 inches.
0
+1
41. yes; x 2 + bx  12
x=1
The solutions are x = 10 and x = 1. However, the diagram
shows that the side length is more than 6 inches, so a side
length of 1 inch does not make sense in this situation.
The original picture, therefore, is 10 inches by 10 inches.
So, the area of the original square picture is 10(10) = 100
square inches.
A =ℓ w
40. a.
( 133 ) ⋅ A
104 = (6 + 7)w
0 = (x  10)(x  1)
+ 10
⋅A
104 = (x + 7)w
0 = x 2  11x + 10
+ 10
Area of
screen
Let A = ℓ w = (x + 7)w and x = 6.
20  20 = x 2  11x + 30  20
Factors of 10
⋅
24 = x(x  2)
24 = x(x) + x(2)
Factors
of −12
Sum of
factors, b
1, 12 1, 12 2, 6 2, 6 3, 4 3, 4
11
11
4
4
1
1
As shown in the table, there are 6 pairs of integer
factors that can be the values of p and q such that
(x + p)(x + q) = x 2 + bx  12. The second row shows
the respective corresponding values of b that go with each
factor pair.
42. a.
24 = x 2  2x
24  24 = x 2  2x  24
0 = x 2  2x  24
0 = (x + 4)(x  6)
3, 3, 4, 4,
2,
2,
1,
Factors 1,
6
8
24 24 12 12
6
8
of −24
Sum of
factors
23
x+4=
4
23
0
4
x = 4
or
10
10
x6=
+6
5
5
2
So, x 2  x  6 = x 2 + (2x  3x)  6 = (x + 2)(x  3).
b.
2
0
+6
x=6
So, x2 + 2x  8 = x2 + (2x + 4x)  8 = (x  2)(x + 4).
The solutions are x = 4 and x = 6, but a length of
4 inches does not make sense in this situation. So, the
length of the browser window is 6 inches.
410
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Chapter 7
A =ℓ w
43.
45. To nd the equation of the form x 2 + bx + c = 0 that
44 = (x  5)(x  12)
44 = x(x) + x(12) + (5)(x) + (5)(12)
44 = x 2  12x  5x + 60
has solutions x = 4 and x = 6, multiply (x + 4)(x  6),
simplify the polynomial, and substitute the values into the
equation.
(x + 4)(x  6) = x(x) + x(6) + 4(x) + 4(6)
44 = x 2  17x + 60
= x 2  6x + 4x  24
44  44 = x 2  17x + 60  44
= x 2  2x  24
0 = x 2  17x + 16
So, the equation x 2  2x  24 = 0 has solutions
x = 4 and x = 6.
0 = (x  1)(x  16)
Factors of 16
1, 16
2, 8
4, 4
17
10
0
Sum of factors
x1=
+1
0 or
+1
x  16 =
+ 16
x=1
0
46. a. Look for the x-intercepts, and substitute them for p and q
in (x  p)(x  q).
b. The graph crosses the x-axis at x = 3 and x = 2.
So, x 2 + x  6 = (x + 3)(x  2).
+ 16
x = 16
The solutions are x = 1 and x = 16. However, a side length
of x  5 = 1  5 = 4 feet does not make sense in this
situation. So, use the solution x = 16. Therefore, the
rectangle is x  5 = 16  5 = 11 feet by
x  12 = 16  12 = 4 feet.
47. a. Area being
=
paved
A =ℓ w +ℓ w
= 20(x) + x(18  x)
= 20x + x(18) + x(x)
A = —12 bh
44.
= 20x + 18x  x 2
35 = —12 (g  8)(g  11)
= x 2 + 38x
35 = —12 [ g(g) + g(11) + (8)(g) + (8)(11) ]
35 = —12 ( g2  11g  8g + 88 )
⋅
2 35 = 2
⋅
(
g2
An expression that represents the area being paved is
( x 2 + 38x ) square meters.
b. A = x 2 + 38x
35 = —12 ( g2  19g + 88 )
1
—2
An equation that can be used to nd the width of the road
is 280 = x 2 + 38x.
 19g + 88 )
280 = x 2 + 38x
70 = g2  19g + 88
280 + x 2 = x 2 + x 2 + 38x
70  70 = g2  19g + 88  70
x 2 + 280 = 38x
0 = g2  19g + 18
x 2 + 280  38x = 38x  38x
0 = (g  1)(g  18)
Factors of 18
1, 18
2, 9
3, 6
19
11
9
Sum of factors
g1=
+1
0
+1
g=1
Area of bottom
Area of top
+
rectangle
rectangle
or
g  18 =
+ 18
0
+ 18
g = 18
The solutions are g = 1 and g = 18. However, a side length
of g  8 = 1  8 = 7 meters does not make sense in this
situation. So, use the solution g = 18. Therefore, the triangle
has a base that is g  8 = 18  8 = 10 meters and a height
that is g  11 = 18  11 = 7 meters.
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x 2  38x + 280 = 0
(x  10)(x  28) = 0
x  10 =
+ 10
0
+ 10
x = 10
or
x  28 =
+ 28
0
+ 28
x = 28
Factors 1, 2, 4, 5, 7, 8, 10, 14,
of 280 280 140 70 56 40 35 28 20
Sum of
factors 281 142 74 61 47 43 38 34
The solutions are x = 10 and x = 28. However, the
diagram shows that the width of the road is less than 18
meters, so a width of 28 meters does not make sense in
this situation. Therefore, the width of the road is
10 meters.
Algebra 1
Worked-Out Solutions
411
Chapter 7
7.6 Explorations (p. 391)
48. x 2 + 6xy + 8y 2
Factors of 8
1. a.
1, 8 2, 4
Sum of factors
9
6
So, x 2 + 6xy + 8y 2 = (x + 2y)(x + 4y).
49. r 2 + 7rs + 12s 2
Factors of 12
So, 3x 2 + 5x + 2 = (x + 1)(3x + 2).
1, 12 2, 6 3, 4
Sum of factors
13
8
b.
7
So, r 2 + 7rs + 12s 2 = (r + 3s)(r + 4s).
50. a 2 + 11ab  26b 2
Factors of −26 1, 26 1, 26 2, 13 2, 13
Sum of factors
25
25
11
11
So, a 2 + 11ab  26b 2 = (a  2b)(a + 13b).
So, 4x 2 + 4x  3 = (2x + 3)(2x  1).
c.
51. x 2  2xy  35y 2
Factors of −35 1, 35 1, 35
Sum of factors
5, 7
5, 7
2
2
34
34
So, x 2  2xy  35y 2 = (x + 5y)(x  7y).
Maintaining Mathematical Prociency
52. p  9 = 0
Check p  9 = 0
?
99=0
+9 +9
p=9
0=0✓
The solution is p = 9.
53. z + 12 = 5
 12
Check
 12
z + 12 = 5
?
17 + 12 = 5
z = 17
5 = 5 ✓
So, 2x 2  11x + 5 = (x  5)(2x  1).
2. Arrange algebra tiles that model the trinomial into a
rectangular array, use additional algebra tiles to model the
dimensions of the rectangle, then write the polynomial in
factored form using the dimensions of the rectangle.
3. The trinomial 2x 2 + 2x + 1 cannot be factored because the
expression cannot be modeled as a rectangular array.
The solution is z = 17.
54.
c
6=—
7
( )
c
7(6) = 7 —
7
42 = c
c
Check 6 = —
7
? 42
6=—
7
6=6✓
The solution is c = 42.
55. 4k = 0
4k 0
—=—
4
4
k=0
7.6 Monitoring Progress (pp. 393–394)
1. 8x 2  56x + 48 = 8( x 2  7x + 6 )
= 8(x  6)(x  1)
Factors of 6
Sum of factors
1, 6 2, 3
7
5
So, 8x 2  56x + 48 = 8(x  6)(x  1).
Check 4k = 0
?
4(0) = 0
0=0✓
The solution is k = 0.
412
Algebra 1
Worked-Out Solutions
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All rights reserved.
Chapter 7
2. 14x 2 + 31x + 15
4. 3x 2  14x + 8
Factors Factors
Possible
of 14
of 15 factorization
1, 14
1, 15
(x + 1)
(14x + 15)
1, 14
15, 1
(x + 15)
(14x + 1)
x + 210x = 211x
1, 14
3, 5
(x + 3)
(14x + 5)
5x + 42x = 47x
1, 14
5, 3
(x + 5)
(14x + 3)
3x + 70x = 73x
2, 7
1, 15
(2x + 1)
(7x + 15)
30x + 7x = 37x
2, 7
15, 1
(2x + 15)
(7x + 1)
2x + 105x = 107x
2, 7
3, 5
(2x + 3)
(7x + 5)
10x + 21x = 31x
(2x + 5)
(7x + 3)
6x + 35x = 41x
2, 7
So,
14x 2
5, 3
Factors Factors
Possible
of 3
of 8
factorization
Middle term
15x + 14x = 29x
✗
✗
✗
✗
✗
✗
✓
1, 3
1, 8
(x  1)
(3x  8)
8x  3x = 11x
1, 3
8, 1
(x  8)
(3x  1)
x  24x = 25x
1, 3
2, 4
(x  2)
(3x  4)
4x  6x = 10x
1, 3
4, 2
(x  4)
(3x  2)
2x  12x = 14x
Middle term
1, 2
1, 5
(x  1)
(2x  5)
5x  2x = 7x
1, 2
5, 1
(x  5)
(2x  1)
x  10x = 11x
So, 2x 2  7x + 5 = (x  1)(2x  5).
✓
✗
✗
✗
✓
5. 4x 2  19x  5
Factors Factors
Possible
of 4
of −5 factorization
Middle term
1, 4
1, 5
(x + 1)
(4x  5)
5x + 4x = x
✗
1, 4
5, 1
(x + 5)
(4x  1)
x + 20x = 19x
✗
1, 4
1, 5
(x  1)
(4x + 5)
5x  4x = x
✗
1, 4
5, 1
(x  5)
(4x + 1)
x  20x = 19x ✓
2, 2
1, 5
(2x + 1)
(2x  5)
10x + 2x = 8x
✗
2, 2
1, 5
(2x  1)
(2x + 5)
10x  2x = 8x
✗
+ 31x + 15 = (2x + 3)(7x + 5).
Factors Factors
Possible
of 2
of 5
factorization
✗
So, 3x 2  14x + 8 = (x  4)(3x  2).
✗
3. 2x 2  7x + 5
Middle term
So, 4x 2  19x  5 = (x  5)(4x + 1).
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Algebra 1
Worked-Out Solutions
413
Chapter 7
6. 6x 2 + x  12
Factors Factors
Possible
of 6
of −12 factorization
Middle term
1, 12
(x + 1)
12x + 6x = 6x
1, 6
✗
(6x  12)
x + 72x = 71x
12, 1
1, 6
(x + 12)
✗
(6x  1)
12x  6x = 6x
1, 12
1, 6
(x  1)
✗
(6x + 12)
x  72x = 71x
12, 1
1, 6
(x  12)
✗
(6x + 1)
6x + 12x = 6x
2, 6
1, 6
(x + 2)
✗
(6x  6)
2x + 36x = 34x
6, 2
1, 6
(x + 6)
✗
(6x  2)
6x  12x = 6x
2, 6
1, 6
(x  2)
✗
(6x + 6)
2x  36x = 34x
6, 2
1, 6
(x  6)
✗
(6x + 2)
4x + 18x = 14x
3, 4
1, 6
(x + 3)
✗
(6x  4)
4, 3
3x + 24x = 21x
1, 6
(x + 4)
✗
(6x  3)
4x  18x = 14x
3, 4
1, 6
(x  3)
✗
(6x + 4)
4, 3
3x  24x = 21x
1, 6
(x  4)
✗
(6x + 3)
24x + 3x = 21x
1, 12
2, 3
(2x + 1)
✗
(3x  12)
12, 1
2, 3
(2x + 12) 2x + 36x = 34x
✗
(3x  1)
24x  3x = 21x
1, 12
2, 3
(2x  1)
✗
(3x + 12)
12, 1
2x  36x = 34x
2, 3
(2x  12)
✗
(3x + 1)
12x + 6x = 6x
2, 6
2, 3
(2x + 2)
✗
(3x  6)
4x + 18x = 14x
6, 2
2, 3
(2x + 6)
✗
(3x  2)
12x  6x = 6x
2, 6
2, 3
(2x  2)
✗
(3x + 6)
6, 2
4x  18x = 14x
2, 3
(2x  6)
✗
(3x + 2)
8x + 9x = x
3, 4
2, 3
(2x + 3)
✓
(3x  4)
6x + 12x = 6x
4, 3
2, 3
(2x + 4)
✗
(3x  3)
8x  9x = x
3, 4
2, 3
(2x  3)
✗
(3x + 4)
4, 3
6x  12x = 6x
2, 3
(2x  4)
✗
(3x + 3)
7. 2y 2  5y  3 = ( 2y 2 + 5y + 3 )
Factors Factors
Possible
of 2
of 3
factorization
Middle term
1, 2
1, 3
( y + 1)
(2y + 3)
3y + 2y = 5y
1, 2
3, 1
( y + 3)
(2y + 1)
y + 6y = 7y
✓
✗
So, 2y 2  5y  3 = ( y + 1)(2y + 3).
8. 5m2 + 6m  1 = ( 5m2  6m + 1 )
Factors Factors
Possible
of 5
of 1
factorization
1, 5
1, 1
(m  1)
(5m  1)
Middle term
m  5m = 6m
✓
So, 5m2 + 6m  1 = (m  1)(5m  1).
9. 3x 2  x + 2 = ( 3x 2 + x  2 )
Factors Factors
Possible
of 3
of −2 factorization
1, 3
1, 2
(x + 1)
(3x  2)
1, 3
2, 1
(x + 2)
(3x  1)
1, 3
1, 2
(x  1)
(3x + 2)
1, 3
2, 1
(x  2)
(3x + 1)
Middle term
2x + 3x = x
x + 6x = 5x
2x  3x = x
x  6x = 5x
✓
✗
✗
✗
So, 3x 2  x + 2 = (x + 1)(3x  2).
So, 6x 2 + x  12 = (2x + 3)(3x  4).
414
Algebra 1
Worked-Out Solutions
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All rights reserved.
Chapter 7
10.
7.6 Exercises (pp. 395–396)
w(2w + 1) = 136
w(2w) + w(1) = 136
Vocabulary and Core Concept Check
2w2 + w = 136
1. The greatest common factor of the terms of
3y 2  21y + 36 is 3.
2w2 + w  136 = 136  136
2w2 + w  136 = 0
2. Factoring 6x 2  x  2 requires considering factors of 6 and
Factors Factors
Possible
of 2 of −136 factorization
Middle term
1, 2
1, 136
(w + 1)
136w + 2w = 134w
✗
(2w  136)
1, 2
136, 1
(w + 136)
(2w  1)
w + 272w = 271w
1, 2
1, 136
(w  1)
(2w + 136)
136w  2w = 134w
1, 2
136, 1
(w  136)
(2w + 1)
1, 2
2, 68
(w + 2)
(2w  68)
w  272w = 271w
68w + 4w = 64w
1, 2
68, 2
(w + 68)
(2w  2)
2w + 136w = 134w
1, 2
2, 68
(w  2)
(2w + 68)
1, 2
68, 2
(w  68)
(2w + 2)
1, 2
4, 34
(w + 4)
(2w  34)
34w + 8w = 26w
1, 2
34, 4
(w + 34)
(2w  4)
4w + 68w = 64w
68w  4w = 64w
2w  136w = 134w
1, 2
4, 34
(w  4)
(2w + 34)
34w  8w = 26w
1, 2
34, 4
(w  34)
(2w + 4)
4w  68w = 64w
1, 2
8, 17
(w + 8)
(2w  17)
17w + 16w = w
1, 2
17, 8
(w + 17)
(2w  8)
8w + 34w = 26w
1, 2
8, 17
(w  8)
(2w + 17)
1, 2
17, 8
(w  17)
(2w + 8)
17w  16w = w
8w  34w = 26w
✗
✗
✗
✗
or
Monitoring Progress and Modeling with Mathematics
3. 3x 2 + 3x  6 = 3( x 2 + x  2 )
= 3(x  1)(x + 2)
Factors of −2
1, 2
1, 2
Sum of factors
1
1
So, 3x 2 + 3x  3 = 3(x  1)(x + 2).
4. 8v 2 + 8v  48 = 8( v 2 + v  6 )
= 8(v  2)(v + 3)
✗
✗
✗
✗
✗
Factors of −6
1, 6
1, 6
2, 3
2, 3
Sum of factors
5
5
1
1
So, 8v 2 + 8v  48 = 8(v  2)(v + 3).
5. 4k 2 + 28k + 48 = 4( k 2 + 7k + 12 )
= 4(k + 3)(k + 4)
Factors of 12
Sum of factors
✗
✗
✗
✓
✗
2w + 17 =
0
17 17
2w = 17
2w 17
—=—
2
2
17
w = —
2
A negative width does not make sense, so you should use the
positive solution. So, the width of the reserve is 8 miles.
Copyright © Big Ideas Learning, LLC
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2, 6
3, 4
13
8
7
So, 4k 2 + 28k + 48 = 4(k + 3)(k + 4).
6. 6y 2  24y + 18 = 6( y 2  4y + 3 )
1, 3
Factors of 3
Sum of factors
4
So, 6y 2  24y + 18 = 6(y  1)(y  3).
7. 7b 2  63b + 140 = 7( b 2  9b + 20 )
= 7(b  4)(b  5)
Factors of 20
Sum of factors
w=8
1, 12
= 6( y  1)(y  3)
✗
(w  8)(2w + 17) = 0
w8= 0
+8 +8
2 in different combinations until the combination is found
that produces the correct middle term. Factoring x 2  x  2
only requires nding the factors of 2 that add up to 1.
1, 20
2, 10
4, 5
22
12
9
So, 7b 2  63b + 140 = 7(b  4)(b  5).
8. 9r 2  36r  45 = 9( r 2  4r  5 )
= 9(r + 1)(r  5)
Factors of −5
1, 5
1, 5
Sum of factors
4
4
So, 9r 2  36r  45 = 9(r + 1)(r  5).
Algebra 1
Worked-Out Solutions
415
Chapter 7
9. 3h2 + 11h + 6
12. 10w2  31w + 15
Factors Factors
Possible
of 3
of 6
factorization
Factors Factors
of 10
of 15
Middle term
1, 3
1, 6
(h + 1)
(3h + 6)
6h + 3h = 9h
1, 3
6, 1
(h + 6)
(3h + 1)
h + 18h = 19h
1, 3
2, 3
(h + 2)
(3h + 3)
3h + 6h = 9h
1, 3
3, 2
(h + 3)
(3h + 2)
2h + 9h = 11h
✗
✗
✗
✓
So, 3h2 + 11h + 6 = (h + 3)(3h + 2).
10. 8m2 + 30m + 7
Factors Factors
Possible
of 8
of 7
factorization
1, 8
1, 7
Middle term
(m + 1)
(8m + 7)
7m + 8m = 15m
1, 8
7, 1
(m + 7)
(8m + 1)
m + 56m = 57m
2, 4
1, 7
(2m + 1)
(4m + 7)
14m + 4m = 18m
2, 4
7, 1
(2m + 7)
(4m + 1)
2m + 28m = 30m
✗
✗
✗
✓
So, 8m2 + 30m + 7 = (2m + 7)(4m + 1).
11. 6x 2  5x + 1
Factors Factors
Possible
of 6
of 1
factorization
Middle term
1, 6
1, 1
(x  1)
(6x  1)
x  6x = 7x
✗
2, 3
1, 1
(2x  1)
(3x  1)
2x  3x = 5x
✓
Possible
factorization
Middle term
1, 10
1, 15
(w  1)
(10w  15)
15w  10w = 25w
1, 10
15, 1
(w  15)
(10w  1)
w  150w = 151w ✗
1, 10
3, 5
(w  3)
(10w  5)
5w  30w = 35w
✗
1, 10
5, 3
(w  5)
(10w  3)
3w  50w = 53w
✗
2, 5
1, 15
(2w  1)
(5w  15)
30w  5w = 35w
✗
2, 5
15, 1
(2w  15)
(5w  1)
2w  75w = 77w
✗
2, 5
3, 5
(2w  3)
(5w  5)
10w  15w = 25w
✗
2, 5
5, 3
(2w  5)
(5w  3)
6w  25w = 31w
✓
✗
So, 10w2  31w + 15 = (2w  5)(5w  3)
13. 3n2 + 5n  2
Factors Factors
Possible
of 3
of −2 factorization
1, 3
1, 2
(n + 1)
(3n  2)
1, 3
2, 1
(n + 2)
(3n  1)
1, 3
1, 2
(n  1)
(3n + 2)
1, 3
2, 1
(n  2)
(3n + 1)
Middle term
2n + 3n = n
n + 6n = 5n
2n  3n = n
n  6n = 5n
✗
✓
✗
✗
So, 3n2 + 5n  2 = (n + 2)(3n  1).
So, 6x 2  5x + 1 = (2x  1)(3x  1).
416
Algebra 1
Worked-Out Solutions
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Chapter 7
14. 4z2 + 4z  3
16. 18v2  15v  18 = 3(6v2  5v  6)
Factors Factors
Possible
of 4
of −3 factorization
Factors Factors
Possible
of 6
of −6 factorization
Middle term
Middle term
1, 4
1, 3
(z + 1)
(4z  3)
3z + 4z = z
✗
1, 6
1, 6
(v + 1)
(6v  6)
6v + 6v = 0
✗
1, 4
3, 1
(z + 3)
(4z  1)
z + 12z = 11z
✗
1, 6
6, 1
(v + 6)
(6v  1)
v + 36v = 35v
✗
1, 4
1, 3
(z  1)
(4z + 3)
3z  4z = z
✗
1, 6
1, 6
(v  1)
(6v + 6)
6v  6v = 0
✗
1, 4
3, 1
(z  3)
(4z + 1)
z  12z = 11z
✗
1, 6
6, 1
(v  6)
(6v + 1)
v  36v = 35v ✗
2, 2
1, 3
(2z + 1)
(2z  3)
6z + 2z = 4z
✗
1, 6
2, 3
(v + 2)
(6v  3)
3v + 12v = 9v
✗
2, 2
1, 3
(2z  1)
(2z + 3)
6z  2z = 4z
✓
1, 6
3, 2
(v + 3)
(6v  2)
2v + 18v = 16v
✗
1, 6
2, 3
(v  2)
(6v + 3)
3v  12v = 9v
1, 6
3, 2
(v  3)
(6v + 2)
2v  18v = 16v ✗
2, 3
1, 6
(2v + 1)
(3v  6)
12v + 3v = 9v
✗
2, 3
6, 1
(2v + 6)
(3v  1)
2v + 18v = 16v
✗
2, 3
1, 6
(2v  1)
(3v + 6)
12v  3v = 9v
✗
2, 3
6, 1
(2v  6)
(3v + 1)
2v  18v = 16v ✗
2, 3
2, 3
(2v + 2)
(3v  3)
6v + 6v = 0
✗
2, 3
3, 2
(2v + 3)
(3v  2)
4v + 9v = 5v
✗
2, 3
2, 3
(2v  2)
(3v + 3)
6v  6v = 0
✗
2, 3
3, 2
(2v  3)
(3v + 2)
4v  9v = 5v
✓
So, 4z2 + 4z  3 = (2z  1)(2z + 3).
15. 8g2  10g  12 = 2(4g2  5g  6)
Factors Factors
Possible
of 4
of −6 factorization
Middle term
1, 4
1, 6
(g + 1)
(4g  6)
6g + 4g = 2g
✗
1, 4
6, 1
(g + 6)
(4g  1)
g + 24g = 23g
✗
1, 4
1, 6
(g  1)
(4g + 6)
6g  4g = 2g
1, 4
6, 1
(g  6)
(4g + 1)
g  24g = 23g ✗
1, 4
2, 3
(g + 2)
(4g  3)
3g + 8g = 5g
2g + 12g = 10g
✗
✗
1, 4
3, 2
(g + 3)
(4g  2)
1, 4
2, 3
(g  2)
(4g + 3)
1, 4
3, 2
(g  3)
(4g + 2)
2g  12g = 10g ✗
2, 2
1, 6
(2g + 1)
(2g  6)
12g + 2g = 10g ✗
2, 2
1, 6
(2g  1)
(2g + 6)
12g  2g = 10g
✗
2, 2
2, 3
(2g + 2)
(2g  3)
6g + 4g = 2g
✗
2, 2
2, 3
(2g  2)
(2g + 3)
6g  4g = 2g
✗
3g  8g = 5g
✗
✓
✗
So, 18v2  15v  18 = 3(2v  3)(3v + 2).
So, 8g2  10g  12 = 2(g  2)(4g + 3).
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Algebra 1
Worked-Out Solutions
417
Chapter 7
20. 8h2  13h + 6 = ( 8h2 + 13h  6 )
17. 3t2 + 11t  6 = (3t2  11t + 6)
Factors Factors
Possible
of 3
of −11 factorization
1, 3
1, 6
(t  1)
(3t  6)
1, 3
6, 1
(t  6)
(3t  1)
Factors Factors
Possible
of 8
of −6 factorization
Middle term
Middle term
6t  3t = 9t
✗
1, 8
1, 6
(h + 1)
(8h  6)
6h + 8h = 2h
✗
t  18t = 19t
✗
1, 8
6, 1
(h + 6)
(8h  1)
h + 48h = 47h
✗
1, 3
2, 3
(t  2)
(3t  3)
3t  6t = 9t
✗
1, 8
1, 6
(h  1)
(8h + 6)
6h  8h = 2h
1, 3
3, 2
(t  3)
(3t  2)
2t  9t = 11t
✓
1, 8
6, 1
(h  6)
(8h + 1)
h  48h = 47h ✗
1, 8
2, 3
(h + 2)
(8h  3)
3h + 16h = 13h
✓
1, 8
3, 2
(h + 3)
(8h  2)
2h + 24h = 22h
✗
1, 8
2, 3
(h  2)
(8h + 3)
3h  16h = 13h ✗
1, 8
3, 2
(h  3)
(8h + 2)
2h  24h = 22h ✗
2, 4
1, 6
(2h + 1)
(4h  6)
12h + 4h = 8h
✗
2, 4
6, 1
(2h + 6)
(4h  1)
2h + 24h = 22h
✗
2, 4
1, 6
(2h  1)
(4h + 6)
12h  4h = 8h
✗
2, 4
6, 1
(2h  6)
(4h + 1)
2h  24h = 22h ✗
2, 4
2, 3
(2h + 2)
(4h  3)
6h + 8h = 2h
✗
2, 4
3, 2
(2h + 3)
(4h  2)
4h + 12h = 8h
✗
2, 4
2, 3
(2h  2)
(4h + 3)
6h  8h = 2h
✗
2, 4
3, 2
(2h  3)
(4h + 2)
4h  12h = 8h
✗
So, 3t2 + 11t  6 = (t  3)(3t  2).
18. 7v2  25v  12 = (7v2 + 25v + 12)
Factors Factors
Possible
of 7
of 12 factorization
1, 7
1, 12
(v + 1)
(7v + 12)
Middle term
12v + 7v = 19v
✗
1, 7
12, 1
(v + 12)
(7v + 1)
v + 84v = 85v
✗
1, 7
2, 6
(v + 2)
(7v + 6)
6v + 14v = 20v
✗
1, 7
6, 2
(v + 6)
(7v + 2)
2v + 42v = 44v
✗
1, 7
3, 4
(v + 3)
(7v + 4)
4v + 21v = 25v
✓
1, 7
4, 3
(v + 4)
(7v + 3)
3v + 28v = 31v
✗
So, 7v2  25v  12 = (v + 3)(7v + 4).
19. 4c2 + 19c + 5 = (4c2  19c  5)
Factors Factors
Possible
of 4
of −5 factorization
Middle term
1, 4
✗
1, 5
(c + 1)
5c + 4c = c
(4c  5)
1, 4
✗
5, 1
(c + 5)
c + 20c = 19c
(4c  1)
1, 4
✗
1, 5
(c  1)
5c  4c = c
(4c + 5)
1, 4
5, 1
(c  5)
c  20c = 19c ✓
(4c + 1)
2, 2
✗
1, 5
(2c + 1)
10c + 2c = 8c
(2c  5)
2, 2
✗
1, 5
(2c  1)
10c  2c = 8c
(2c + 5)
✗
So, 8h2  13h + 6 = (h + 2)(8h  3).
So, 4c2 + 19c + 5 = (c  5)(4c + 1).
418
Algebra 1
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Chapter 7
21. 15w2  w + 28 = (15w2 + w  28)
Factors Factors
Possible
of 15
of −28 factorization
Middle term
1, 15
1, 28
(w + 1)
28w + 15w = 13w
(15w  28)
1, 15
28, 1
(w + 28)
w + 420w = 419w
(15w  1)
1, 15
1, 28
(w  1)
28w  15w = 13w
(15w + 28)
1, 15
28, 1
(w  28)
w  420w = 419w
(15w + 1)
1, 15
2, 14
(w + 2)
14w + 30w = 16w
(15w  14)
1, 15
14, 2
(w + 14)
2w + 210w = 208w
(15w  2)
1, 15
2, 14
(w  2)
14w  30w = 16w
(15w + 14)
1, 15
14, 2
(w  14)
2w  210w = 208w
(15w + 2)
1, 15
4, 7
(w + 4)
7w + 60w = 53w
(15w  7)
1, 15
7, 4
(w + 7)
4w + 105w = 101w
(15w  4)
1, 15
4, 7
(w  4)
7w  60w = 53w
(15w + 7)
1, 15
7, 4
(w  7)
4w  105w = 101w
(15w + 4)
3, 5
1, 28
(3w + 1)
84w + 5w = 79w
(5w  28)
3, 5
28, 1
(3w + 28) 3w + 140w = 137w
(5w  1)
3, 5
1, 28
(3w  1)
84w  5w = 79w
(5w + 28)
3, 5
28, 1
(3w  28)
3w  140w = 137w
(5w + 1)
3, 5
2, 14
(3w + 2)
42w + 10w = 32w
(5w  14)
3, 5
14, 2
(3w + 14)
6w + 70w = 64w
(5w  2)
3, 5
2, 14
(3w  2)
42w  10w = 32w
(5w + 14)
3, 5
14, 2
(3w  14)
6w  70w = 64w
(5w + 2)
3, 5
4, 7
(3w + 4)
21w + 20w = w
(5w  7)
3, 5
7, 4
(3w + 7)
12w + 35w = 23w
(5w  4)
3, 5
4, 7
(3w  4)
21w  20w = w
(5w + 7)
3, 5
7, 4
(3w  7)
12w  35w = 23w
(5w + 4)
22. 22d 2 + 29d  9 = (22d 2  29d + 9)
Factors Factors
Possible
of 22
of 9
factorization
Middle term
✗
1, 22
1, 9
(d  1)
(22d  9)
9d  22d = 31d
✗
1, 22
9, 1
(d  9)
(22d  1)
d  198d = 199d ✗
✗
1, 22
3, 3
(d  3)
(22d  3)
3d  66d = 69d
✗
✗
2, 11
1, 9
18d  11d = 29d
✓
✗
(2d  1)
(11d  9)
2, 11
9, 1
2d  99d = 101d ✗
✗
(2d  9)
(11d  1)
2, 11
3, 3
(2d  3)
(11d  3)
6d  33d = 39d
✗
✗
✗
✗
✗
✗
✗
✗
So, 22d 2 + 29d  9 = (2d  1)(11d  9).
23. The greatest common factor 2 should be factored out of
every term, but it was only factored out of the rst term.
2x 2  2x  24 = 2(x 2  x  12)
= 2(x + 3)(x  4)
Factors 1, 12 1, 12
of −12
Sum of
11
11
factors
2, 6
2, 6
3, 4
3, 4
4
4
1
1
So, 2x 2  2x  24 = 2(x + 3)(x  4).
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
✓
✗
So, 15w2  w + 28 = (3w  4)(5w + 7).
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Algebra 1
Worked-Out Solutions
419
Chapter 7
24. These factors do not produce the correct middle term when
26. 2k2  5k  18 = 0
they are multiplied.
Factors Factors
Possible
of 2
of −18 factorization
6x2  7x  3
Factors Factors
Possible
of 6
of −3 factorization
(k + 1)
(2k  18)
18k + 2k = 16k ✗
1, 3
(x + 1)
(6x  3)
3x + 6x = 3x
✗
1, 2
18, 1
(k + 18)
(2k  1)
k + 36k = 35k
✗
1, 6
3, 1
(x + 3)
(6x  1)
x + 18x = 17x
✗
1, 2
1, 18
(k  1)
(2k + 18)
18k  2k = 16k
✗
1, 6
1, 3
(x  1)
(6x + 3)
3x  6x = 3x
✗
1, 2
18, 1
(k  18)
(2k + 1)
1, 6
3, 1
(x  3)
(6x + 1)
x  18x = 17x
✗
1, 2
2, 9
(k + 2)
(2k  9)
2, 3
1, 3
(2x + 1)
(3x  3)
6x + 3x = 3x
✗
1, 2
9, 2
(k + 9)
(2k  2)
2, 3
3, 1
(2x + 3)
(3x  1)
2x + 9x = 7x
✗
1, 2
2, 9
(k  2)
(2k + 9)
2, 3
1, 3
(2x  1)
(3x + 3)
6x  3x = 3x
✗
1, 2
9, 2
(k  9)
(2k + 2)
2, 3
3, 1
(2x  3)
(3x + 1)
2x  9x = 7x
✓
1, 2
3, 6
(k + 3)
(2k  6)
6k + 6k = 0
✗
1, 2
6, 3
(k + 6)
(2k  3)
3k + 12k = 9k
✗
1, 2
3, 6
(k  3)
(2k + 6)
1, 2
6, 3
(k  6)
(2k + 3)
5x 2  5x  30 = 0
25.
5(x 2
⋅
 x  6) = 0
5(x + 2)(x  3) = 0
1
5(x + 2)(x  3) = —
5
(x + 2)(x  3) = 0
x+2=
2
0
or
⋅0
2
k  36k = 35k ✗
9k + 4k = 5k
2k + 18k = 16k
9k  4k = 5k
✓
✗
✗
2k  18k = 16k ✗
6k  6k = 0
3k  12k = 9k
✗
✗
(k + 2)(2k  9) = 0
x3=
+3
x = 2
k+2=
0
2
+3
Factors of −6
1, 6
1, 6
2, 3
2, 3
Sum of factors
5
5
1
1
Algebra 1
Worked-Out Solutions
0
2
or
2k  9 =
+9
0
+9
k = 2
x=3
The roots are x = 2 and x = 3.
420
1, 18
1, 6
So, 6x 2  7x  3 = (2x  3)(3x + 1).
1
—
5
1, 2
Middle term
Middle term
2k = 9
2k 9
—=—
2
2
9
k=—
2
9
The roots are k = 2 and k = —.
2
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Chapter 7
27.
12n2  11n = 15
(3n + 5)(4n  3) = 0
12n2  11n + 15 = 15 + 15
1[ (3n + 5)(4n  3) ] = 1(0)
12n2  11n + 15 = 0
(3n + 5)(4n  3) = 0
(12n2 + 11n  15) = 0
Factors Factors
Possible
of 12
of −15 factorization
Middle term
1, 15
(n + 1)
15n + 12n = 3n
1, 12
(12n  15)
15, 1
(n + 15)
n + 180n = 179n
1, 12
(12n  1)
1, 15
(n  1)
15n  12n = 3n
1, 12
(12n + 15)
15, 1
(n  15)
n  180n = 179n
1, 12
(12n + 1)
3, 5
(n + 3)
5n + 36n = 31n
1, 12
(12n  5)
5, 3
(n + 5)
3n + 60n = 57n
1, 12
(12n  3)
3, 5
(n  3)
5n  36n = 31n
1, 12
(12n + 5)
5, 3
(n  5)
3n  60n = 57n
1, 12
(12n + 3)
1, 15
(2n + 1)
30n + 6n = 24n
2, 6
(6n  15)
15, 1
(2n + 15)
2n + 90n = 88n
2, 6
(6n  1)
1, 15
(2n  1)
30n  6n = 24n
2, 6
(6n + 15)
15, 1
(2n  15)
2n  90n = 88n
2, 6
(6n + 1)
3, 5
(2n + 3)
10n + 18n = 3n
2, 6
(6n  5)
5, 3
(2n + 5)
6n + 30n = 24n
2, 6
(6n  3)
3, 5
(2n  3)
10n  18n = 8n
2, 6
(6n + 5)
5, 3
(2n  5)
6n  30n = 24n
2, 6
(6n + 3)
1, 15
(3n + 1)
45n + 4n = 41n
3, 4
(4n  15)
15, 1
(3n + 15)
3n + 60n = 57n
3, 4
(4n  1)
1, 15
(3n  1)
45n  4n = 41n
3, 4
(4n + 15)
15, 1
(3n  15)
3n  60n = 57n
3, 4
(4n + 1)
3, 5
(3n + 3)
15n + 12n = 3n
3, 4
(4n  5)
5, 3
(3n + 5)
9n + 20n = 11n
3, 4
(4n  3)
3, 5
(3n  3)
15n  12n = 3n
3, 4
(4n + 5)
5, 3
(3n  5)
9n  20n = 11n
3, 4
(4n + 3)
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3n + 5 =
5
✗
✗
5
+3
0
+3
4n = 3
4n 3
—=—
4
4
3
n=—
4
5
3
The roots are n = — and n = —.
3
4
✗
✗
4n  3 =
or
3n = 5
3n 5
—=—
3
3
5
n = —
3
✗
✗
0
14b 2  2 = 3b
28.
14b 2
 2 + 3b = 3b + 3b
14b 2
+ 3b  2 = 0
Factors Factors
Possible
of 14
of −2 factorization
Middle term
1, 14
1, 2
(b + 1)
(14b  2)
2b + 14b = 12b
✗
✗
1, 14
2, 1
(b + 2)
(14b  1)
b + 28b = 27b
✗
✗
1, 14
1, 2
(b  1)
(14b + 2)
2b  14b = 12b ✗
✗
1, 14
2, 1
(b  2)
(14b + 1)
b  28b = 27b ✗
✗
2, 7
1, 2
4b + 7b = 3b
✓
✗
(2b + 1)
(7b  2)
2, 7
2, 1
(2b + 2)
(7b  1)
2b + 14b = 12b
✗
2, 7
1, 2
(2b  1)
(7b + 2)
2, 7
2, 1
(2b  2)
(7b + 1)
✗
✗
✗
✗
✗
✗
✗
✗
✗
4b  7b = 3b
✗
2b  14b = 12b ✗
(2b + 1)(7b  2) = 0
2b + 1 = 0
1 1
or
7b  2 = 0
+2 +2
2b = 1
2b 1
—=—
2
2
1
b = —
2
7b = 2
7b 2
—=—
7
7
2
b=—
7
1
2
The roots are b = — and b = —.
2
7
✗
✓
✗
✗
Algebra 1
Worked-Out Solutions
421
Chapter 7
29. Let y = 0.
30. Let y = 0.
y = 2x2  3x  35
y = 4x 2 + 11x  3
0=
0 = 4x 2 + 11x  3
2x2
 3x  35
Factors Factors
Possible
of 2
of −35 factorization
Factors Factors
Possible
of 4
of −3 factorization
Middle term
1, 2
1, 35
(x + 1)
(2x  35)
35x + 2x = 33x ✗
1, 2
35, 1
(x + 35)
(2x  1)
x + 70x = 69x
1, 2
1, 35
(x  1)
(2x + 35)
35x  2x = 33x
1, 2
35, 1
(x  35)
(2x + 1)
1, 2
5, 7
(x + 5)
(2x  7)
7x + 10x = 3x
1, 2
7, 5
(x + 7)
(2x  5)
5x + 14x = 9x
1, 2
5, 7
(x  5)
(2x + 7)
7x  10x = 3x
✓
1, 2
7, 5
(x  7)
(2x + 5)
5x  14x = 9x
✗
1, 4
1, 3
(x + 1)
(4x  3)
3x + 4x = x
✗
✗
1, 4
3, 1
(x + 3)
(4x  1)
x + 12x = 11x
✓
✗
1, 4
1, 3
(x  1)
(4x + 3)
3x  4x = x
✗
x  70x = 69x ✗
1, 4
3, 1
(x  3)
(4x + 1)
x  12x = 11x
✗
✗
2, 2
1, 3
(2x + 1)
(2x  3)
6x + 2x = 4x
✗
✗
2, 2
1, 3
(2x  1)
(2x + 3)
6x  2x = 4x
✗
0 = (x  5)(2x + 7)
x5= 0
+5 +5
or
2x + 7 = 0
7 7
x=5
2x = 7
2x 7
—=—
2
2
7
x = —
2
The x-coordinates of the points where the graph crosses the
7
x-axis are the roots x = 5 and x = —.
2
422
Algebra 1
Worked-Out Solutions
Middle term
0 = (x + 3)(4x  1)
x+3=
3
0
3
or
4x  1 =
+1
0
+1
x = 3
4x = 1
4x 1
—=—
4
4
1
x=—
4
The x-coordinates of the points where the graph crosses the
1
x-axis are the roots x = 3 and x = —.
4
Copyright © Big Ideas Learning, LLC
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Chapter 7
31. Let y = 0.
32. Let y = 0.
y = 7x 2  2x + 5
0=
7x 2
y = 3x 2 + 14x + 5
 2x + 5
0 = 3x 2 + 14x + 5
0 = (7x 2 + 2x  5)
1(0) = 1[
(7x 2
0 = (3x 2  14x  5)
+ 2x  5) ]
1(0) = 1[ (3x 2  14x  5) ]
0 = 7x 2 + 2x  5
0 = 3x 2  14x  5
Factors Factors
Possible
of 7
of −5 factorization
Factors Factors
Possible
of 3
of −5 factorization
Middle term
1, 7
1, 5
(x + 1)
(7x  5)
5x + 7x = 2x
1, 7
5, 1
(x + 5)
(7x  1)
x + 35x = 34x
1, 7
1, 5
(x  1)
(7x + 5)
1, 7
5, 1
(x  5)
(7x + 1)
✓
1, 5
(x + 1)
(3x  5)
5x + 3x = 2x
✗
✗
1, 3
5, 1
(x + 5)
(3x  1)
x + 15x = 14x
✗
5x  7x = 2x
✗
1, 3
1, 5
(x  1)
(3x + 5)
5x  3x = 2x
✗
x  35x = 34x
✗
1, 3
5, 1
(x  5)
(3x + 1)
x  15x = 14x
✓
0 = (x + 1)(7x  5)
x+1=
1
0
1
or
Middle term
1, 3
0 = (x  5)(3x + 1)
7x  5 =
+5
x = 1
0
+5
7x = 5
7x 5
—=—
7
7
5
x=—
7
The x-coordinates of the points where the graph crosses the
5
x-axis are the roots x = 1 and x = —.
7
x5=
+5
0
or
+5
3x + 1 =
1
0
1
x=5
3x = 1
3x 1
—=—
3
3
1
x = —
3
The x-coordinates of the points where the graph crosses the
1
x-axis are the roots x = 5 and x = —.
3
33. a. 15x 2  x  2 = (3x + 1)(?)
Factors Factors
of 15
of −2 Factorization
3, 5
1, 2
(3x + 1)
(5x  2)
Middle term
6x + 5x = x
✓
So, 15x 2  x  2 = (3x + 1)(5x  2) and the length of
the sign is represented by (5x  2) feet.
b. One way to nd the area of the sign when x = 3 is to
substitute 3 into the expression for the area 15x2  x  2
and then simplify. Another way is to substitute 3 into the
expressions for the length (5x  2) and width (3x + 1). Then
simplify each expression and multiply these two values.
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Algebra 1
Worked-Out Solutions
423
Chapter 7
34. Let h = 0.
Factors of
−1085
h = 16t2 + 8t + 80
0 = 16t2 + 8t + 80
w  31 =
+ 31
Middle term
(t + 1)
(2t  10)
10t + 2t = 8t
✗
1, 2
10, 1
(t + 10)
(2t  1)
t + 20t = 19t
✗
1, 2
1, 10
(t  1)
(2t + 10)
10t  2t = 8t
✗
1, 2
10, 1
(t  10)
(2t + 1)
1, 2
2, 5
(t + 2)
(2t  5)
1, 2
5, 2
(t + 5)
(2t  2)
1, 2
2, 5
(t  2)
(2t + 5)
1, 2
5, 2
(t  5)
(2t + 2)
t  20t = 19t
5t + 4t = t
✗
or
7, 155
7, 155
31, 35
31, 35
1084
1084
212
212
148
148
4
4
w + 35 =
0
+ 31
 35
w = 31
w = 35
36. Let w be the width of the invitation, then the length is 2w  1.
A =ℓ w
15 = (2w  1)w
✗
15  15 = 2w 2  w  15
2t + 10t = 8t
0
 35
A negative width does not make sense, so you should use the
positive solution. So, the width of the base is 31 meters, and
the length is 8 + 2(31) = 70 meters.
✓
15 = 2w 2  w
0 = 2w 2  w  15
5t  4t = t
2t  10t = 8t
2t  5 = 0
+5 +5
t = 2
5, 217
15 = 2w(w)  1(w)
0 = (t + 2)(2t  5)
t+2= 0
2 2
5, 217
0 = (w  31)(w + 35)
Factors Factors
Possible
of 2
of −10 factorization
1, 10
1, 1085
Sum of
factors
0 = 8(2t2  t  10)
1
1
— (0) = — [ 8(2t2  t  10) ]
8
8
0 = 2t2  t  10
1, 2
1, 1085
2t = 5
2t 5
—=—
2
2
t = 2.5
A negative time does not make sense, so you should use the
positive solution. So, the diver is in the air for 2.5 seconds.
✗
✗
Factors Factors
Possible
of 2
of 15 factorization
1, 2
1, 15
(w + 1)
(2w  15)
1, 2
15, 1
(w + 15)
(2w  1)
w + 30w = 29w
✗
1, 2
1, 15
(w  1)
(2w + 15)
15w  2w = 13w
✗
1, 2
15, 1
(w  15)
(2w + 1)
1, 2
3, 5
(w + 3)
(2w  5)
5w + 6w = w
✗
1, 2
5, 3
(w + 5)
(2w  3)
3w + 30w = 27w
✗
1, 2
3, 5
(w  3)
(2w + 5)
5w  6w = w
✓
1, 2
5, 3
(w  5)
(2w + 3)
3w  10w = 7w
✗
35. Let w be the width of the base, then the length is 8 + 2w.
A =ℓ w
2170 = (8 + 2w)(w)
2170 = 8(w) + 2w(w)
Middle term
15w + 2w = 13w ✗
w  30w = 29w ✗
2170 = 8w + 2w2
2170  2170 = 8w + 2w2  2170
0 = 2w2 + 8w  2170
0 = 2(w2 + 4w  1085)
1
—2 (0)
= —12 [ 2(w2 + 4w  1085) ]
0 = w2 + 4w  1085
424
Algebra 1
Worked-Out Solutions
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Chapter 7
0 = (w  3) (2w + 5)
w3=
+3
0
+3
or
41. 2x 2 + tx + 10
2w + 5 =
5
0
5
w=3
2w = 5
2w 5
—=—
2
2
5
w = —
2
A negative width does not make sense, so you should use the
positive solution. So, the width of the invitation is 3 inches,
and the length is 2(3)  1 = 6  1 = 5 inches. The invitation
will t inside the envelope because the width is less than
1
5
3 — inches, and the length is less than 5 — inches.
8
8
37. Sample answer: 3x(x + 2) = 3x(x) + 3x(2)
=
3x2
+ 6x
So, one binomial that has a GCF of 3x is 3x2 + 6x.
38. The graph of g represents function k, and the graph of h
represents function ℓ . Because c is positive in each function,
the constant terms in the factors must have the same sign.
Because g has a positive value of b, the constant terms of the
factors will both be positive, which results in negative roots,
and the graph of k has two negative x-intercepts. Because h
has a negative value of b, the constant terms of the factors
will both be negative, which results in positive roots, and the
graph of ℓ has two positive x-intercepts.
Factors Factors
Possible
of 2
of 10 factorization
Middle term
1, 2
1, 10
(x + 1)
(2x + 10)
10x + 2x = 12x
1, 2
1,
10
(x  1)
(2x  10)
10x  2x = 12x
1, 2
10, 1
(x + 10)
(2x + 1)
x + 20x = 21x
1, 2
10,
1
(x  10)
(2x  1)
x  20x = 21x
1, 2
2, 5
(x + 2)
(2x + 5)
5x + 4x = 9x
1, 2
2, 5
(x  2)
(2x  5)
5x  4x = 9x
1, 2
5, 2
(x + 5)
(2x + 2)
2x + 10x = 12x
1, 2
5, 2
(x  5)
(2x  2)
2x  10x = 12x
So, the values of t that make 2x 2 + tx + 10 factorable are
±9, ±12, and ±21.
42 a.
39. It is not possible to factor ax 2 + bx + c, where a ≠ 1,
when no combination of factors of a and c produce the
correct middle term. Sample answer: One such trinomial
is 5x 2  x + 3.
40. Your friend is incorrect. To use the Zero-Product Property,
2x 2 + 5x  3 = (x + 3)(2x  1)
b.
one side of the equation needs to be 0. So, you must rst
subtract 2 from each side of the equation, then factor.
3x 2  2x  1 = (x  1)(3x + 1)
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Algebra 1
Worked-Out Solutions
425
Chapter 7
43. Let w be the width (in inches) of the rectangle, then the
44. Let x be the width (in feet) of the patio.
length is 2w + 1.
ℓ w = 5 + 2ℓ + 2w
Area of
Area of pool
Area of pool
=

pool surface
surface and patio
surface
(2w + 1)w = 5 + 2(2w + 1) + 2w
2w(w) + 1(w) = 5 + 2(2w) + 2(1) + 2w
(16)(24) = (x + 24 + x)(x + 16 + x)  (16)(24)
2w 2 + w = 5 + 4w + 2 + 2w
2w 2
Area of
patio
Area of
=
pool surface
A=5+P
384 = (2x + 24)(2x + 16)  384
+ w = 4w + 2w + 5 + 2
384 = 2x(2x) + 2x(16) + 24(2x) + 24(16)  384
2w 2 + w = 6w + 7
+ w  6w = 6w  6w + 7
384 = 4x 2 + 32x + 48x + 384  384
2w 2  5w = 7
384 = 4x 2 + 80x
2w 2  5w  7 = 7  7
384  384 = 4x 2 + 80x  384
2w 2
0 = 4x 2 + 80x  384
2w 2  5w  7 = 0
Factors Factors
Possible
of 2
of −7 factorization
0 = 4( x 2 + 20x  96 )
1
1, 2
1, 7
(w + 1)
(2w  7)
7w + 2w = 5w
✓
1, 2
7, 1
(w + 7)
(2w  1)
w + 14w = 13w
✗
1, 2
1, 7
(w  1)
(2w + 7)
7w  2w = 5w
✗
1, 2
7, 1
(w  7)
(2w + 1)
w  14w = 13w ✗
1
0
1
w = 1
or
2w  7 =
+7
0
+7
2w = 7
2w 7
—=—
2
2
w = 3.5
A negative width does not make sense, so you should use the
positive solution. So, the width of the rectangle is 3.5 inches.
1
]
0 = x 2 + 20x  96
Factors
of −96
Sum of
factors
(w + 1)(2w  7) = 0
w+1=
[
—4 (0) = —4 4( x 2 + 20x  96 )
Middle term
1,
96
1,
96
2,
48
2,
48
3,
32
3,
32
4,
24
4,
24
6,
16
6,
16
8,
12
8,
12
95
95
46
46
29
29
20
20
10
10
4
4
0 = (x  4)(x + 24)
x4=
+4
x=
0
+4
4
or
x + 24 =
 24
0
 24
x = 24
A negative width does not make sense, so you should use the
positive solution. So, the width of the patio is 4 feet.
45. 4k 2 + 7jk  2j 2
Factors Factors
Possible
of 4
of −2 factorization
Middle term
1, 4
1, 2
(k + j)
(4k  2j)
2jk + 4jk = 2jk
✗
1, 4
2, 1
(k + 2j)
(4k  j)
jk + 8jk = 7jk
✓
1, 4
1, 2
(k  j)
(4k + 2j)
2jk  4jk = 2jk
✗
1, 4
2, 1
(k  2j)
(4k + j)
jk  8jk = 7jk
✗
2, 2
1, 2
(2k + j)
(2k  2j)
4jk + 2jk = 2jk
✗
2, 2
1, 2
(2k  j)
(2k + 2j)
4jk  2jk = 2jk
✗
So, 4k 2 + 7jk  2j 2 = (k + 2j)(4k  j).
426
Algebra 1
Worked-Out Solutions
Copyright © Big Ideas Learning, LLC
All rights reserved.
Chapter 7
48. 18m 3 + 39m 2n  15mn 2 = 3m( 6m 2 + 13mn  5n 2 )
46. 6x 2 + 5xy  4y 2
Factors Factors
Possible
of 6
of −4 factorization
Factors Factors
Possible
of 6
of −5 factorization
Middle term
Middle term
1, 6
1, 4
(x + y)
(6x  4y)
4xy + 6xy = 2xy
✗
1, 6
1, 5
(m + n)
(6m  5n)
5mn + 6mn = mn
✗
1, 6
4, 1
(x + 4y)
(6x  y)
xy + 24xy = 23xy
✗
1, 6
5, 1
(m + 5n)
(6m  n)
mn + 30mn = 29mn
✗
1, 6
1, 4
(x  y)
(6x + 4y)
4xy  6xy = 2xy
✗
1, 6
1, 5
(m  n)
(6m + 5n)
5mn  6mn = mn
✗
1, 6
4, 1
(x  4y)
(6x + y)
xy  24xy = 23xy ✗
1, 6
5, 1
(m  5n)
(6m + n)
mn  30mn = 29mn ✗
1, 6
2, 2
(x + 2y)
(6x  2y)
✗
2, 3
1, 5
(2m + n)
(3m  5n)
10mn + 3mn = 7mn
✗
1, 6
2, 2
(x  2y)
(6x + 2y)
2xy  12xy = 10xy ✗
2, 3
5, 1
(2m + 5n)
(3m  n)
2mn + 15mn = 13mn
✓
2, 3
1, 4
(2x + y)
(3x  4y)
✗
2, 3
1, 5
(2m  n)
(3m + 5n)
10mn  3mn = 7mn
✗
2, 3
4, 1
(2x + 4y)
(3x  y)
✗
2, 3
5, 1
(2m  5n)
(3m + n)
2mn  15mn = 13mn ✗
2, 3
1, 4
(2x  y)
(3x + 4y)
2, 3
4, 1
(2x  4y)
(3x + y)
2, 3
2, 2
(2x + 2y)
(3x  2y)
2, 3
2, 2
(2x  2y)
(3x + 2y)
2xy + 12xy = 10xy
8xy + 3xy = 5xy
2xy + 12xy = 10xy
8xy  3xy = 5xy
✓
2xy  12xy = 10xy ✗
4xy + 6xy = 2xy
4xy  6xy = 2xy
✗
✗
47. 6a 2 + 19ab  14b 2 =  ( 6a 2  19ab + 14b 2 )
Middle term
1,
14
(a  b)
(6a  14b)
14ab  6ab = 20ab ✗
1, 6
14,
1
(a  14b)
(6a  b)
ab  84ab = 85ab ✗
1, 6
2, 7
(a  2b)
(6a  7b)
7ab  12ab = 19ab ✓
1, 6
7, 2
(a  7b)
(6a  2b)
2ab  42ab = 44ab ✗
2, 3
1,
14
(2a  b)
(3a  14b)
28ab  3ab = 31ab ✗
2, 3
14,
1
(2a  14b)
(3a  b)
2ab  42ab = 44ab ✗
2, 3
2, 7
(2a  2b)
(3a  7b)
14ab  6ab = 20ab ✗
(2a  7b)
(3a  2b)
4ab  21ab = 25ab ✗
7, 2
⋅
—
49. ± √ 64 = ± √ 8 8 = ± 8
—
⋅
—
50. √ 4 = √ 2 2 = 2
⋅
52. ± √ 81 = ± √ 9⋅9 = ± 9
—
—
51. √ 225 = √ 15 15 = 15
—
—
53. y = 3 + 7x
y  x = 3
Step 1 Equation 1 is already solved for y.
Step 2
y  x = 3
(3 + 7x)  x = 3
1, 6
2, 3
Maintaining Mathematical Prociency
—
So, 6x 2 + 5xy  4y 2 = (2x  y)(3x + 4y).
Factors Factors
Possible
of 6
of 14 factorization
So, 18m 3 + 39m 2n  15mn 2 = 3m (2m + 5n)(3m  n).
So, 6a 2 + 19ab  14b 2 = (a  2b)(6a  7b).
7x  x + 3 = 3
6x + 3 = 3
3
3
6x = 6
6x 6
—=—
6
6
x = 1
Step 3 y = 3 + 7x
y = 3 + 7(1)
y=37
y = 4
Check
y = 3 + 7x
?
4 = 3 + 7(1)
?
4 = 3  7
4 = 4 ✓
The solution is (1, 4).
Copyright © Big Ideas Learning, LLC
All rights reserved.
y  x = 3
?
4  (1) = 3
?
4 + 1 = 3
3 = 3 ✓
Algebra 1
Worked-Out Solutions
427
Chapter 7
56. x  8 = y
54. 2x = y + 2
9y  12 + 3x = 0
x + 3y = 14
Step 1
2x = y + 2
Step 1
2x  2 = y + 2  2
2x  2 = y
x + 3y = 14
Step 2
Step 2
x + 3(2x  2) = 14
9y  12 + 3x = 0
9(x + 8) 12 + 3x = 0
x + 3(2x)  3(2) = 14
9(x) + 9(8)  12 + 3x = 0
x + 6x  6 = 14
9x + 72  12 + 3x = 0
5x  6 = 14
+6
x  8 = y
x  8 y
—=—
1
1
x+8=y
12x + 60 =
+6
 60
5x = 20
5x 20
—=—
5
5
x=4
Step 3
12x = 60
12x 60
—=—
2
12
x = 5
2x = y + 2
Step 3
2(4) = y + 2
5  8 = y
2
3 = y
3 y
—=—
1 1
3=y
6=y
Check
x  8 = y
(5)  8 = y
8=y+2
2
2x = y + 2
?
2(4) = 6 + 2
x + 3y = 14
?
4 + 3(6) = 14
?
4 + 18 = 14
8=8✓
14 = 14 ✓
The solution is (4, 6).
Check
x  8 = y
?
(5)  8 = 3
?
5  8 = 3
3 = 3 ✓
55. 5x  2y = 14
9y  12 + 3x = 0
?
9(3)  12 + 3(5) = 0
?
27  12  15 = 0
?
15  15 = 0
0=0✓
7 = 2x + y
Step 1
0
 60
The solution is (5, 3).
7 = 2x + y
7 + 2x = 2x + 2x + y
2x  7 = y
5x  2y = 14
Step 2
5x  2(2x  7) = 14
5x  2(2x)  2(7) = 14
5x  4x + 14 = 14
x + 14 =
 14
Step 3 7 = 2x + y
14
 14
x=0
7 = 2(0) + y
7 = 0 + y
7 = y
Check
5x  2y = 14
?
5(0)  2(7) = 14
?
0 + 14 = 14
14 = 14 ✓
7 = 2x + y
?
7 = 2(0) + (7)
?
7 = 0  7
7 = 7 ✓
The solution is (0, 7).
428
Algebra 1
Worked-Out Solutions
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Chapter 7
7.7 Explorations (p. 397)
2.
1. a.
4x 2  1 = 4x 2  2x + 2x  1 = (2x + 1)(2x  1)
This product is a sum and difference pattern.
b.
9x 2  6x + 6x  4 = 9x 2  4 = (3x + 2)(3x  2)
4x 2  4x + 1 = (2x  1)(2x  1) = (2x  1)2
This product is a square of a binomial pattern.
c.
9x 2  6x  6x + 4 = 9x 2  12x + 4 = (3x  2)2
4x 2 + 4x + 1 = (2x + 1)(2x + 1) = (2x + 1)2
This product is a square of a binomial pattern.
d.
9x 2 + 6x + 6x + 4 = 9x 2 + 12x + 4 = (3x + 2)2
3. Sample answer: The algebra tiles for special patterns will
4x 2  6x + 2 = (2x  2)(2x  1)
This product is not a special pattern.
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always form a square array, and the x 2 tiles will also form
a square array. Factor special products by using the special
patterns studied in Lesson 7.3, but in reversed order: For the
sum and difference pattern, the number of x and x tiles will
be the same, and there will only be 1 tiles to complete the
array. For the square of a binomial pattern, the x tiles will
either be all positive or all negative and there will only be +1
tiles to complete the array.
Algebra 1
Worked-Out Solutions
429
Chapter 7
4. a.
+
+
+
+
+
+
+
+
+
+
+
+
2. 100  m 2 = 10 2  m 2
= (10 + m)(10  m)
So, 100  m 2 = (10 + m)(10  m).
+
3. 9n 2  16 = (3n)2  42
+
+
+
+
+
+
+
= (3n + 4)(3n  4)
So, 9n 2  16 = (3n + 4)(3n  4).
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
5. 362  342 = (36 + 34)(36  34)
+
+
+
+
+
+
+
= 70(2)
4. 16h 2  49 = (4h)2  72
= (4h + 7)(4h  7)
So, 16h 2  49 = (4h + 7)(4h  7).
= 140
25x 2 + 10x + 1 = (5x + 1)2
b.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
 342 = 140.
6. 472  442 = (47 + 44)(47  44)
= 91(3)
= 273
So, 472  442 = 273.
−
+
+
−
−
+
So,
362
7. 552  502 = (55 + 50)(55  50)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
−
−
−
−
−
−
−
+
= 105(5)
−
+
= 525
−
So,
+
+
+
 502 = 525.
8. 282  242 = (28 + 24)(28  24)
= 52(4)
= 208
So, 282  242 = 208.
25x 2  10x + 1 = (5x  1)2
c.
552
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
9. m 2  2m + 1 = m 2  2(m)(1) + 12
= (m  1)2
So,
m2
 2m + 1 = (m  1)2.
10. d 2  10d + 25 = d 2  2(d )(5) + 52
= (d  5)2
So,
d2
 10d + 25 = (d  5)2.
11. 9z 2 + 36z + 36 = 9( z 2 + 4z + 4 )
= 9[ z 2 + 2(z)(2) + 22 ]
= 9(z + 2)2
+
+
+
+
+
+
+
−
−
−
−
−
−
−
( 25x 2  1 ) = (5x  1)(5x + 1)
7.7 Monitoring Progress (pp. 398–400)
1.
x2
 36 =
x2

62
= (x + 6)(x  6)
So, x 2  36 = (x + 6)(x  6).
430
Algebra 1
Worked-Out Solutions
So, 9z 2 + 36z + 36 = 9(z + 2)2.
a 2 + 6a + 9 = 0
12.
a2
+ 2(a)(3) + 32 = 0
(a + 3)2 = 0
a+3=
3
0
3
a = 3
The equation has a repeated root of a = 3.
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All rights reserved.
Chapter 7
49
7
w 2  —w + — = 0
3
36
13.
(
Monitoring Progress and Modeling with Mathematics
3. m 2  49 = m 2  72
)
( )
49
7
36 w 2  —w + — = 36(0)
3
36
7
49
2
36( w )  36 —w + 36 — = 0
3
36
( )
= (m + 7)(m  7)
So,
= (z + 9)(z  9)
(6w)2  2(6w)(7) + 72 = 0
(6w 
+7
So,
=0
6w  7 =
14.
z2
 81 = (z + 9)(z  9).
5. 64  81d 2 = 82  (9d)2
0
= (8 + 9d )(8  9d)
So, 64  81d 2 = (8 + 9d)(8  9d ).
+7
6w = 7
7
6w
—=—
6
6
7
w=—
6
7
The equation has a repeated root of w = — .
6
n2
6. 25  4x 2 = 52  (2x)2
= (5 + 2x)(5  2x)
So, 25  4x 2 = (5 + 2x)(5  2x).
7. 225a 2  36b 2 = 9( 25a 2  4b 2 )
= 9[ (5a)2  (2b)2 ]
 81 = 0
= 9(5a + 2b)(5a  2b)
n 2  92 = 0
So, 225a 2  36b 2 = 9(5a + 2b)(5a  2b).
(n + 9)(n  9) = 0
n+9=
9
0
n9=
or
9
+9
n = 9
0
+9
n=9
8. 16x 2  169y 2 = (4x)2  (13y)2
= (4x + 13y)(4x  13y)
So, 16x 2  169y 2 = (4x + 13y)(4x  13y).
9. 122  92 = (12 + 9)(12  9)
The roots are n = 9 and n = 9.
= 21(3)
15. Let y = 0.
= 63
y = 81  16t 2
So,
0 = 81  16t 2
122
= 30(8)
0 = (9 + 4t)(9  4t)
9 + 4t =
 92 = 63.
10. 192  112 = (19 + 11)(19  11)
0 = 92  (4t)2
9
 49 = (m + 7)(m  7).
4. z 2  81 = z 2  92
36w 2  84w + 49 = 0
7)2
m2
0
9
= 240
9  4t = 0
or
9
4t = 9
4t 9
—=—
4
4
t = 2.25
9
4t = 9
4t 9
—=—
4
4
t = 2.25
A negative time does not make sense in this situation. So, the
golf ball hits the ground after 2.25 seconds.
So, 192  112 = 240.
11. 782  722 = (78 + 72)(78  72)
= 150(6)
= 900
So,
782
12. 542  522 = (54 + 52)(54  52)
= 106(2)
7.7 Exercises (pp. 401–402)
Vocabulary and Core Concept Check
1. The square roots of the rst and last terms are y and 8, and
⋅ ⋅
the middle term is 2 y 8, so it ts the perfect square
trinomial pattern and can be factored as such.
= 212
So, 542  522 = 212.
13. 532  472 = (53 + 47)(53  47)
= 100(6)
2. The polynomial that does not belong is k 2 + 25, because it is
the only one that cannot be factored using a special pattern.
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 722 = 900.
= 600
So,
532
 472 = 600.
Algebra 1
Worked-Out Solutions
431
Chapter 7
14. 392  362 = (39 + 36)(39  36)
25. a. d 2 + 8d + 16 = d 2 + 2(d)(4) + 42
= 75(3)
= (d + 4)2
= 225
So, an expression that represents the side length of the
coaster is (d + 4) centimeters.
So, 392  362 = 225.
b. P = 4(s)
15. h 2 + 12h + 36 = h 2 + 2(h)(6) + 62
= 4(d + 4)
= (h + 6)2
= 4(d) + 4(4)
So, h 2 + 12h + 36 = (h + 6)2.
= 4d + 16
An expression for the perimeter of the coaster is
(4d + 16) centimeters.
16. p 2 + 30p + 225 = p 2 + 2( p)(15) + 152
= ( p + 15)2
So, p 2 + 30p + 225 = ( p + 15)2.
26. a. A = x 2  30x + 225
= x 2  2(x)(15) + 152
17. y 2  22y + 121 = y 2  2( y)(11) + 112
= (x  15)2
= (y  11)2
A polynomial that represents the side length of the
playground is (x  15) feet.
So, y 2  22y + 121 = (y  11)2.
18. x 2  4x + 4 = x 2  2(x)(2) + 22
b. P = 4(s)
= (x  2)2
= 4(x  15)
So, x 2  4x + 4 = (x  2)2.
= 4(x)  4(15)
19. a 2  28a + 196 = a2  2(a)(14) + 142
= 4x  60
= (a  14)2
An expression for the perimeter of the playground is
(4x  60) feet.
So, a2  28a + 196 = (a  14)2.
20. m 2 + 24m + 144 = m2 + 2(m)(12) + 122
27.
z 2  22 = 0
= (m + 12)2
(z + 2)(z  2) = 0
So, m 2 + 24m + 144 = (m + 12)2.
z+2=
21. 25n 2 + 20n + 4 = (5n)2 + 2(5n)(2) + 22
= (5n +
z2  4 = 0
2
2)2
23. The difference of two squares pattern should be used to
factor this polynomial.
n 2  64 = n 2  82
= (n + 8)(n  8)
So,
n2
 64 = (n + 8)(n  8).
24. The perfect square trinomial pattern should be used to factor
this polynomial.
y 2  6y + 9 = y 2  2(y)(3) + 32
= (y  3)2
So,
432
y2
 6y + 9 = ( y  3)2.
Algebra 1
Worked-Out Solutions
z2=
+2
0
+2
z=2
The roots are z = 2 and z = 2.
22. 49a 2  14a + 1 = (7a)2  2(7a)(1) + 12
So, 49a 2  14a + 1 = (7a  1)2.
or
z = 2
So, 25n 2 + 20n + 4 = (5n + 2)2.
= (7a  1)2
0
2
4x 2 = 49
28.
4x 2
 49 = 49  49
4x 2  49 = 0
(2x)2  72 = 0
(2x + 7)(2x  7) = 0
2x + 7 =
7
0
or
7
2x  7 =
+7
2x = 7
2x 7
—=—
2
2
7
x=—
2
7
7
The roots are x =  — and x = —.
2
2
0
+7
2x = 7
2x 7
—=—
2
2
7
x=—
2
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Chapter 7
29.
k 2  16k + 64 = 0
33.
k 2  2(k)(8) + 82 = 0
(k  8)2 = 0
k8=
+8
0
+8
k=8
The equation has a repeated root of k = 8.
30.
s 2 + 20s + 100 = 0
1
1
y2 + —y = —
16
2
1
1
1
1
y2 + —y + — = — + —
16 16
2
16
1
1
y2 + —y + — = 0
2
16
1
1
16 y2 + —y + — = 16(0)
2
16
1
1
16( y2 ) + 16 —y + 16 — = 0
2
16
16y2 + 8y + 1 = 0
(
( )
(4y)2 + 2(4y)(1) + 12 = 0
s 2 + 2(s)(10) + 10 2 = 0
(4y + 1)2 = 0
(s + 10) 2 = 0
s + 10 =
 10
n2 + 9 = 6n
n2 + 9  6n = 6n  6n
n2  6n + 9 = 0
n2  2(n)(3) + 32 = 0
(n  3)2 = 0
n3=
0
+3
n=3
The equation has a repeated root of n = 3.
32.
1
 10
The equation has a repeated root of s = 10.
+3
4y + 1 =
0
s = 10
31.
y 2 = 12y  36
y 2  12y = 12y  12y  36
y 2  12y = 36
4y = 1
(
(3x  2)2 = 0
3x  2 =
+2
(y  6)2 = 0
+6
)
( ) ()
(3x)2  2(3x)(2) + 22 = 0
y 2  12y + 36 = 0
y 2  2(y)(6) + 62 = 0
0
+6
y=6
The equation has a repeated root of y = 6.
0
1
4y 1
—=—
4
4
1
y = —
4
1
The equation has a repeated root of y = —.
4
4
4
2
34.
—x + — = x
3
9
4
4
—x + — + x2 = x2 + x2
3
9
4
4
x2  —x + — = 0
3
9
4
4
2
9 x  —x + — = 9(0)
3
9
4
4
2
9( x )  9 —x + 9 — = 0
3
9
9x2  12x + 4 = 0
y 2  12y + 36 = 36 + 36
y6=
)
( )
0
+2
3x = 2
3x 2
—=—
3
3
2
x=—
3
2
The equation has a repeated root of x = —.
3
35. 3z2  27 = 3( z2  9 )
= 3( z2  32 )
= 3(z + 3)(z  3)
So, 3z2  27 = 3(z + 3)(z  3).
36. 2m2  50 = 2( m2  25 )
= 2( m2  52 )
= 2(m + 5)(m  5)
So,
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2m2
 50 = 2(m + 5)(m  5).
Algebra 1
Worked-Out Solutions
433
Chapter 7
37. 4y 2  16y + 16 = 4( y 2  4y + 4 )
42. Let y = 1.
= 4[ y 2  2(y)(2) + 22 ]
y = 16t 2 + 8t
= 4( y 
1 = 16t 2 + 8t
2)2
So, 4y 2  16y + 16 = 4( y  2)2.
1 + 16t 2 = 16t 2 + 16t + 8t
8k 2 + 80k + 200 = 8(k 2 + 10k + 25)
38.
= 8[ k + 2(k)(5) + 52 ]
= 8(k + 5)2
39.
16t 2  8t + 1 = 0
(4t  1)2 = 0
+ 120y + 72 =
+ 60y + 36)
= 2[ (5y)2 + 2(5y)(6) + 62 ]
2(25y 2
= 2(5y + 6)2
So, 50y 2 + 120y + 72 = 2(5y + 6)2.
40. 27m2  36m + 12 = 3(9m2  12m + 4)
= 3[ (3m)2  2(3m)(2) + 22 ]
= 3(3m  2)2
So, 27m2  36m + 12 = 3(3m  2)2.
41. Let y = 0.
4t  1 =
+1
Factors of 84
0 = (5 + 4t)(5  4t)
Sum of factors
0
5
+1
4t = 1
4t 1
—=—
4
4
1
t = —, or 0.25
4
1
The equation has a repeated root of t = —. So, the
4
grasshopper is 1 foot off the ground 0.25 second after the
start of the jump.
0 = 25  16t 2
5 + 4t =
0
43. a. w 2 + 18w + 84
y = 25  16t 2
5
16t 2 + 1  8t = 8t  8t
(4t)2  2(4t)(1) + 12 = 0
So, 8k 2 + 80k + 200 = 8(k + 5)2.
50y 2
16t 2 + 1 = 8t
or
5  4t =
5
0
5
4t = 5
4t = 5
4t 5
4t 5
—=—
—=—
4
4
4
4
5
5
t = —, or 1.25
t = —, or 1.25
4
4
A negative time does not make sense in this situation. So, the
paintbrush lands on the ground 1.25 seconds after you drop it.
1, 84 2, 42 3, 28 4, 21 6, 14 7, 12
85
44
31
25
20
19
The trinomial w2 + 18w + 84 cannot be factored because
no factor pair of 84 has a sum of 18. However, if you
change the value of c to the perfect square 81, the
trinomial w2 + 18w + 81 can be factored using the perfect
square trinomial pattern.
w2 + 18w + 81 = ( w2 + 2(w)(9) + 92 )
= (w + 9)2
So, w2 + 18w + 81 can be factored, but w2 + 18w + 84
cannot.
b. y2  10y + 23
Factors of 23
1, 23
Sum of factors
24
The trinomial y 2  10y + 23 cannot be factored because
the only factor pair of 23 does not have a sum of 10.
However, if you change the value of c to the perfect
square 25, the trinomial y2  10y + 25 can be factored
using the perfect square trinomial pattern.
y 2  10y + 25 = y 2  2(y)(5) + 52
= (y  5)2
So, y 2  10y + 25 can be factored, but y 2  10y + 23
cannot.
434
Algebra 1
Worked-Out Solutions
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Chapter 7
44. a.
b. Set 9x2  144 equal to 81 and solve.
9x2  144 = 81
9x2
 144  81 = 81  81
9x2  225 = 0
9( x2  25 ) = 0
—9 [ 9( x2  25 ) ] = —9 (0)
1
1
4x2  1 = (2x + 1)(2x  1)
x2  25 = 0
b.
x2  52 = 0
(x + 5)(x  5) = 0
x+5=
5
x5=
or
+5
x = 5
4x2  4x + 1 = (2x  1)2
0
+5
x=5
The side length cannot be negative. So, the solution is
x = 5, which means that the side length of one picture
frame is 5 inches.
45. To simplify (2x  5)2  (x  4)2, you can square each
binomial and then combine like terms. Or, you can use the
difference of two squares pattern with each binomial as
one of the terms. Sample answer: Choose the difference
of two squares pattern so that you do not need to square
any binomials.
0
5
48. a.
Volume of the
Volume
Volume of other
=
+
composite solid
of cube
rectangular prism
= s 3 + ℓ wh
= (x)3 + 4(x)(4)
46. a. When you subtract b 2 from a 2, you “take away” the purple
= x 3 + 16x
region. So, the remaining blue, yellow, and red regions
represent a 2  b 2, as shown below.
A polynomial that represents the volume of the composite
solid is ( x 3 + 16x ) cubic inches.
a+b
Yellow
Red
Blue
b. Set 25x equal to x 3 + 16x and solve.
25x = x 3 + 16x
ab
25x  25x = x 3 + 16x  25x
0 = x 3  9x
0 = x( x 2  9 )
b. (a  b)2 is represented by the yellow region. a 2 is
0 = x( x 2  32 )
the entire square. Subtracting 2ab requires removing
2 regions that each represent ab, which would be the blue
and purple regions, and the red and purple regions. There
is only one purple region, and subtracting 2ab required
removing it twice, so one purple region needs to be
replaced, which adds b 2.
47. a. 9
⋅
0 = x(x + 3)(x  3)
x = 0 or x + 3 =
3
= 9( x2  16 )
=
9( x2 )
=
9x2
x3=
+3
x = 3
0
+3
x=3
Maintaining Mathematical Prociency
49.
50
⋅
⋅ ⋅
10 5
 9(16)
 144
A polynomial that represents the area of the picture
frames, not including the pictures, is ( 9x2  144 )
square inches.
or
Side lengths cannot be negative or 0. So, the solution is
x = 3.
Area of
Area of frame
Area of
=9

frame
and picture
picture
= 9( x2  42 )
0
3
2 5 5
⋅ ⋅
⋅
The prime factorization of 50 is 2 5 5, or 2 52.
50.
44
⋅
⋅ ⋅
4 11
2 2 11
⋅ ⋅
⋅
The prime factorization of 44 is 2 2 11, or 22 11.
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Algebra 1
Worked-Out Solutions
435
Chapter 7
51.
85
3y + 3 < x
56.
⋅
3y + 3  3 < x  3
5 17
The prime factorization of 85 is 5 17.
⋅
3y < x  3
x3
3
1
y < —x  1
3
Now the inequality is in slope-intercept form. So, use the
1
1
slope m = — and the y-intercept b = 1 to graph y = — x  1.
3
3
Because the symbol is <, use a dashed line, and shade below it.
3y
3
96
52.
—<—
⋅ 12
2 ⋅4 ⋅ 4 ⋅3
2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅3
8
⋅ ⋅ ⋅ ⋅ ⋅
⋅
The prime factorization of 96 is 2 2 2 2 2 3, or 25 3.
53. Because the inequality is in slope-intercept form, use
the slope m = 4, and the y-intercept b = 1 to graph
y = 4x  1. Because the symbol is ≤ , use a solid line, and
shade below it.
y
4
2
y
4
4
2
4
4 x
2
6
2
2
4 x
7.8 Explorations (p. 403)
1. a. (x + 1)(x + 1)(2)
54. Because the inequality is in slope-intercept form, use
the slope m =  —12 and the y-intercept b = 3 to graph
y =  —12x + 3. Because the symbol is >, use a dashed line,
and shade above it.
8
+
+
+
+
+
+
+
+
So, (x + 1)(x + 1)(2) = ( x2 + 2x + 1 )(2)
y
= x2(2) + 2x(2) + 1(2)
6
= 2x2  4x  2.
b. (x + 2)(x + 1)(x)
4
+
+
+
+
+
4y  12 ≥ 8x
+
+
+
+
+
4y  12 + 12 ≥ 8x + 12
2
x
2
55.
2
4
6
4y ≥ 8x + 12
4y 8x + 12
—≥—
4
4
y ≥ 2x + 3
Now the inequality is in slope-intercept form. So, use the
slope m = 2 and the y-intercept b = 3 to graph y = 2x + 3.
Because the symbol is ≥, use a solid line, and shade above it.
6
+
So, (x + 2)(x + 1)(x) = ( x2 + 3x + 2 )(x)
= x2(x) + 3x(x) + 2(x)
= x3  3x2  2x.
y
4
4
2
4 x
2
436
Algebra 1
Worked-Out Solutions
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Chapter 7
c. (x + 3)(x)(2)
2. a. J; x3 + x2 = x2(x + 1)
b. A; x3  x = x( x2  1 )
+
+
+
+
+
= x( x2  12 )
= x(x + 1)(x  1)
+
+
+
+
d. G;
x3

4x2
+ 4x = x( x2  4x + 4 )
= x( x2  2(x)(2) + 22 )
= x2(2) + 3x(2)
= 2x2 + 6x
d. (x + 1)(x  1)(x)
= x(x  2)2
e. N; x3  2x2  3x = x( x2  2x  3 )
= x(x + 1)(x  3)
f. B; x3  2x2 + x = x( x2  2x + 1 )
= x( x  2(x)(1) + 1 )
−
+
+
−
+
+
−
= x(x  1)2
g. F;
+ (x  x)  1 ](x)
= x2(x)  1(x)
= x3  x.
e. (x + 1)(x + 1)(x)
+
−
−
−
+
+
+
= x(x  1)(x  2)
k. M; x3 + 2x2  3x = x( x2 + 2x  3 )
= x(x  1)(x  3)
n. H; x3 + 4x2 + 4x = x( x2 + 4x + 4 )
= x3  x.
f. (x  1)(x + 1)(2)
= x( x2 + 2(x)(2) + 22 )
= x(x + 2)2
o. C; x3 + 2x2 + x = x( x2 + 2x + 1 )
= x( x + 2(x)(1) + 12 )
= x(x + 1)2
Factor out the greatest common monomial factor rst, then
factor the remaining expression if possible.
+
−
−
−
j. E; x3  3x2 + 2x = x( x2  3x + 2 )
m. K; x3  2x2 = x2(x  2)
= x2(x) + 1(x)
−
i. I; x3  x2 = x2(x  1)
= x(x + 3)(x  1)
= ( x2 + 1 )(x)
−
h. L; x3 + 2x2 = x2(x + 2)
l. O; x3  4x2 + 3x = x( x2  4x + 3 )
So, (x + 1)(x + 1)(x) = [ x2 + (x  x) + 1 ](x)
−
 4x = x( x2  4 )
= x(x + 2)(x  2)
x2
= ( x2  1 )(x)
+
x3
= x( x2  22 )
So, (x + 1)(x  1)(x) = [
+
+
x2
= x(x + 2)(x  1)
So, (x + 3)(x)(2) = ( x2 + 3x )(2)
+
 2x = x( x2 + x  2 )
c. D;
x3
3. Factor out the greatest common monomial factor rst, then
factor the remaining expression if possible.
So, (x  1)(x + 1)(2) = ( x2  2x  1 )(2)
= x2(2)  2x(2)  1(2)
= 2x2+ 4x + 2.
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Algebra 1
Worked-Out Solutions
437
Chapter 7
4. a. x3 + 4x2 + 3x = x( x2 + 4x + 3 )
x( x2  25 ) = 0
= x(x + 1)(x + 3)
+ 3x = x(x + 1)(x + 3).
b. x3  6x2 + 9x = x( x2  6x + 9 )
x3
So,
+
4x2
= x(
x2
 2(x)(3) +
x3  25x = 0
7.
32
)
x( x2  52 ) = 0
x(x + 5)(x  5) = 0
x=0
or
x+5=
= x(x  3)2
5
So, x3  6x2 + 9x = x(x  3) 2.
c. x3 + 6x2 + 9x = x( x2 + 6x + 9 )
= x( x2 + 2(x)(3) + 32 )
= x(x + 3)2
So, x3 + 6x2 + 9x = x(x + 3)2.
7.8 Monitoring Progress (pp. 404–406)
1.
a3
+
3a2
+a+3=
=
( a3
+
a 2(a
So,
+
3a2
3)(a2
+ 1)
+ a + 3 = (a + 3)( a2 + 1 ).
2. y2 + 2x + yx + 2y = y2 + yx + 2x + 2y
= y( y + x) + 2(x + y)
= y(x + y) + 2(x + y)
= (x + y)( y + 2).
So, y2 + 2x + yx + 2y = (x + y)( y + 2).
3. 3x3  12x = 3x( x2  4 )
= 3x(x + 2)(x  2)
So, 3x3  12x = 3x(x + 2)(x  2).
4.

12y2
+ 18y = 2y(
y2
 6y + 9 )
= 2y( y  2( y)(3) + 32 )
= 2y( y 
3)2
So, 2y3  12y2 + 18y = 2y( y  3)2.
5. m3  2m2  8m = m( m2  2m  8 )
= m(m  4)(m + 2)
So,
6.
m3

2m2
 8m = m(m  4)(m + 2).
w3  8w2 + 16w = 0
w( w2  8w + 16 ) = 0
w(w  4)2 = 0
or
w4=
+4
0
+4
w=4
The equation has a root of w = 0 and a repeated root of
w = 4.
438
x = 5
0
+5
x=5
8. c3  7c2 + 12c = 0
c( c2  7c + 12 ) = 0
c(c  3)(c  4) = 0
c=0
or
c3=
+3
0
c4=
or
+3
+4
c=3
0
+4
c=4
The roots are c = 0, c = 3, and c = 4.
9. Volume = length
72 =
x
⋅
⋅
⋅
⋅
width
(x  1)
72 = x(x  1)(x + 9)
height
(x + 9)
72 = x[ x(x) + x(9)  1(x)  1(9) ]
72 = x( x2 + 9x  x  9 )
72 = x( x2 + 8x  9 )
72 = x(x)2 + x(8x)  x(9)
72  72 = x3 + 8x2  9x  72
0 = x3 + 8x2  9x  72
0 = ( x3 + 8x2 ) + (9x  72)
0 = x2(x + 8)  9(x + 8)
0 = (x + 8)( x2  9 )
0 = (x + 8)( x2  32 )
0 = (x + 8)(x + 3)(x  3)
x+8=
8
0
8
x = 8
or
x+3=
3
0
or
3
x = 3
x3=
+3
0
+3
x=3
Disregard x = 8 and x = 3 because x represents the
length of the box, and a negative length does not make sense.
So, 3 is the correct value of x.
Use x = 3 to nd the width and height, as shown.
w( w2  2(w)(4) + 42 ) = 0
w=0
+5
72 = x3 + 8x2  9x
= 3x( x2  22 )
2y3
x5=
or
The roots are x = 0, x = 5, and x = 5.
+ (a + 3)
+ 3) + (a + 3)
= (a +
a3
3a2 )
0
5
Algebra 1
Worked-Out Solutions
width = x  1
height = x + 9
=31
=3+9
=2
= 12
The length is 3 feet, the width is 2 feet, and the height is
12 feet.
Copyright © Big Ideas Learning, LLC
All rights reserved.
Chapter 7
11. 2x3  2x = 2x( x2  1 )
7.8 Exercises (pp. 407–408)
= 2x( x2  12 )
Vocabulary and Core Concept Check
= 2x(x + 1)(x  1)
1. It is written as a product of unfactorable polynomials with
So,
integer coefcients.
2. Look for terms with common factors.
2x3
 2x = 2x(x + 1)(x  1).
12. 36a 4  4a 2 = 4a 2( 9a 2  1 )
= 4a 2( (3a)2  12 )
Monitoring Progress and Modeling with Mathematics
3. x3 + x2 + 2x + 2 = ( x3 + x2 ) + (2x + 2)
= x2(x + 1) + 2(x + 1)
= (x + 1)( x2 + 2 )
So,
x3
+
x2
+ 2x + 2 = (x + 1)( x2 + 2 ).
4. y3  9y2 + y  9 = ( y3  9y2 ) + ( y  9)
5.

So,
y3
9y2
3z3
+ 2z 
So, 36a 4  4a 2 = 4a 2(3a + 1)(3a  1).
13. 2c2  7c + 19
Factors Factors
of 2
of 19
Possible
factorization
1, 2
1, 19 (c  1)(2c  19) 19c  2c = 21c ✗
= ( y  9)(
1, 2
19, 1 (c  19)(2c  1) c  38c = 39c ✗
y2
+ 1)
8=
3z3

12z2
+ 2z  8
= ( 3z3  12z2 ) + ( 2z  8 )
= 3z2(z  4) + 2(z  4)
= (z  4)( 3z2 + 2 )
So, 3z3 + 2z  12z2  8 = (z  4)( 3z2 + 2 ).
6. 2s3  27  18s + 3s2 = 2s3 + 3s2  18s  27
= ( 2s3 + 3s2 ) + (18s  27)
= s2(2s + 3)  9(2s + 3)
= (2s + 3)( s2  9 )
The polynomial 2c2  7c + 19 is unfactorable because
no combination of factors of 2 and 19 produce a middle
term of 7c.
14. m2  5m  35
Factors of −35
1, 35
1, 35
5, 7
5, 7
Sum of factors
−34
34
−2
2
The polynomial m2  5m  35 is unfactorable because
a = 1, and no factor pairs of 35 have a sum of 5.
15. 6g3  24g2 + 24g = 6g( g2  4g + 4 )
= 6g( g2  2(g)(2) + 2 2 )
= (2s + 3)( s2  32 )
= (2s + 3)(s + 3)(s  3)
So, 2s3  27  18s + 3s2 = (2s + 3)(s + 3)(s  3).
7. x2 + xy + 8x + 8y = ( x2 + xy ) + (8x + 8y)
= x(x + y) + 8(x + y)
= (x + y)(x + 8)
So,
x2
+ xy + 8x + 8y = (x + y)(x + 8).
8. q2 + q + 5pq + 5p = ( q2 + q ) + (5pq + 5p)
= q(q + 1) + 5p(q + 1)
= (q + 1)(q + 5p)
So,
q2
Middle term
= y2( y  9) + ( y  9)
+ y  9 = ( y  9)( y2 + 1 ).
12z2
= 4a 2(3a + 1)(3a  1)
= 6g(g  2) 2
So,
6g3

24g2
+ 24g = 6g(g  2)2.
16. 15d 3 + 21d 2  6d = 3d( 5d 2  7d + 2 )
= 3d(d  1)(5d  2)
So,
15d 3
+
21d 2
17. 3r 5 + 3r 4  90r 3 = 3r 3( r 2 + r  30 )
= 3r 3(r  5)(r + 6)
So,
3r 5
+
3r 4
 90r 3 = 3r 3(r  5)(r + 6).
18. 5w4  40w3 + 80w2 = 5w2( w2  8w + 16 )
+ q + 5pq + 5p = (q + 1)(q + 5p).
9. m2  3m + mn  3n = ( m2  3m ) + (mn  3n)
= m(m  3) + n(m  3)
= (m  3)(m + n)
So, m2  3m + mn  3n = (m  3)(m + n).
10. 2a 2 + 8ab  3a  12b = ( 2a 2 + 8ab ) + (3a  12b)
= 2a(a + 4b)  3(a + 4b)
= (a + 4b)(2a  3)
So, 2a 2 + 8ab  3a  12b = (a + 4b)(2a  3).
Copyright © Big Ideas Learning, LLC
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 6d = 3d(d  1)(5d  2).
= 5w2( w2  2(w)(4) + 4 2 )
= 5w2(w  4) 2
So, 5w4  40w3 + 80w2 = 5w2(w  4) 2.
19. 4c4 + 8c3  28c2 = 4c2( c2  2c + 7 )
Factors of 7
Sum of factors
1, 7
8
c2  2c + 7 is unfactorable, so the polynomial is factored
completely.
So, 4c4 + 8c3  28c2 = 4c2( c2  2c + 7 ).
Algebra 1
Worked-Out Solutions
439
Chapter 7
20. 8t2 + 8t  72 = 8( t2 + t  9 )
Factors of −9
x3 + x2 = 4x + 4
25.
x3 + x2  4x = 4x  4x + 4
1, 9 1, 9 3, 3
Sum of factors
8
x3 + x2  4x = 4
0
8
x3 + x2  4x  4 = 4  4
t2 + t  9 is unfactorable, so the polynomial is factored
completely.
So, 8t2 + 8t  72 = 8( t2 + t  9 ).
x3 + x2  4x  4 = 0
( x3 + x2 ) + (4x  4) = 0
x2(x + 1)  4(x + 1) = 0
(x + 1)( x2  4 ) = 0
21. b3  5b2  4b + 20 = ( b3  5b2 ) + (4b + 20)
=
b2(b
(x + 1)( x2  2 2 ) = 0
 5)  4(b  5)
= (b  5)( b2  4 )
= (b 
5)( b2

(x + 1)(x + 2)(x  2) = 0
22 )
x+1=
1
= (b  5)(b + 2)(b  2)
So,
b3

5b2
2t3( t2 + t  72 ) = 0
= (h + 4)( h2  5 2 )
2t3(t + 9)(t  8) = 0
4)( h2
 25h  100 = (h + 4)(h + 5)(h  5).
3—
 6n + 8 ) = 0
√t3
n2=
0
or n  4 =
+2 +2
+4
n=2
+4
27.
n=4
k 2(k + 10)(k  10) = 0
0
k 2 = √0
k=0
or
k + 10 =
 10
3—
t8=0
+8
+8
t=8
= √0
12s  3s3 = 0
3s(2 + 5)(2  5) = 0
k 2( k 2  10 2 ) = 0
—
x=2
3s( 2 2  s2 ) = 0
k 2( k 2  100 ) = 0
k2 =
+2
3s( 4  s2 ) = 0
k 4  100k 2 = 0
—
+2
The equation has roots t = 9, t = 8, and a repeated root
t = 0.
0
The roots are n = 0, n = 2, and n = 4.
24.
x2= 0
t=0
5n(n  2)(n  4) = 0
5n = 0 or
5n 0
—=—
5
5
n=0
or
x = 2
2t3 = 0 or t + 9 = 0 or
2t3 0
9 9
—=—
2
2
t3 = 0
t = 9
23. 5n3  30n2 + 40n = 0
5n( n2
0
2
The roots are x = 1, x = 2, and x = 2.
= (h + 4)(h + 5)(h  5)
+
2
 25 )
= (h +
So,
x+2=
26. 2t5 + 2t4  144t3 = 0
= h2(h + 4)  25(h + 4)
4h2
or
x = 1
 4b + 20 = (b  5)(b + 2)(b  2).
22. h3 + 4h2  25h  100 = ( h3 + 4h2 ) + (25h  100)
h3
0
1
0
 10
k = 10
or
k  10 =
0
+ 10 + 10
3s = 0 or 2 + s = 0
3s 0
2
2
—=—
3
3
s=0
s = 2
or
2s= 0
+s
+s
2=s
The roots are s = 0, s = 2, and s = 2.
k = 10
The equation has roots of k = 10, k = 10, and repeated
roots of k = 0.
440
Algebra 1
Worked-Out Solutions
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Chapter 7
4y3  7y2 + 28 = 16y
28.
31. Let y = 0.
4y3  7y2 + 28  16y = 16y  16y

y =2x4 + 16x3  32x2
 16y + 28 = 0
0 = 2x4 + 16x3  32x2
( 4y3  7y2 ) + (16y + 28) = 0
0 = 2x2( x2  8x + 16 )
4y3
y2(4y
7y2
 7)  4(4y  7) = 0
0 = 2x2( x2  2(x)(4) + 4 2 )
(4y  7)( y2  4 ) = 0
0 = 2x2(x  4) 2
(4y  7)(
y2

22
)=0
(4y  7)( y + 2)( y  2) = 0
4y  7 =
+7
0
y+2=
or
+7
2
4y = 7
4y 7
—=—
4
4
7
y=—
4
0
y2=
or
2
+2
y = 2
0
+2
y=2
= √0
x=0
0 = x( 4x2 + 25x  56 )
0 = x(x + 8)(4x  7)
x=0
 81 )
or
x+8=
8
0 = x( x2  9 2 )
or x + 9 =
9
0
x9= 0
or
9
+9
x = 9
+9
x=9
The x-coordinates of the points where the graph crosses the
x-axis are the roots x = 0, x = 9, and x = 9.
30. Let y = 0.
4x  7 =
+7
0
+7
4x = 7
4x 7
—=—
4
4
7
x=—
4
The x-coordinates of the points where the graph crosses the
7
x-axis are the roots x = 0, x = 8, and x = —.
4
a 3 + 8a 2  6a  48 = ( a 3 + 8a 2 ) + (6a  48)
0 = 3x4  24x3  45x2
= a 2(a + 8)  6(a + 8)
0 = 3x2( x2 + 8x + 15 )
= (a + 8)( a 2  6 )
0 = 3x2(x + 3)(x + 5)
—
8
or
33. In the second group, factor out 6 instead of 6.
y = 3x4  24x3  45x2
or
0
x = 8
0 = x(x + 9)(x  9)
—
x=4
—
0 = 4x3 + 25x2  56x
0 = x3  81x
3x2 = 0
0
3x2
—=—
3
3
x2 = 0
+4
y = 4x3 + 25x2  56x
y = x3  81x
x=0
√ x2
+4
0
32. Let y = 0.
29. Let y = 0.
0=
—
x4=
The x-coordinates of the points where the graph crosses the
x-axis are the repeated roots x = 0 and x = 4.
7
The roots are y = —, y = 2, and y = 2.
4
x( x2
2x2 = 0 or
0
2x2
—=—
2
2
x2 = 0
x+3=
3
0
3
x = 3
or
x+5=
5
0
5
x = 5
√ x2 = √ 0
x=0
The x-coordinates of the points where the graph crosses
the x-axis are the roots x = 5, x = 3, and the repeated
root x = 0.
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So,
a3
+
8a 2
 6a  48 = (a + 8)( a 2  6 ).
34. It is not factored completely because x2  9 can be factored.
x3  6x2  9x + 54 = ( x3  6x2 ) + (9x + 54)
= x2(x  6)  9(x  6)
= (x  6)( x2  9 )
= (x  6)( x2  3 2 )
= (x  6)(x + 3)(x  3)
So, x3  6x2  9x + 54 = (x  6)(x + 3)(x  3).
Algebra 1
Worked-Out Solutions
441
Chapter 7
35. a. Volume = length
=
4
⋅
⋅
⋅ height
⋅ (w + 4)
width
w
= 4w(w + 4)
Volume =
36.
⋅
⋅
⋅
⋅
width
w
1152 = w(2w + 4)(18  w)
height
(18  w)
= 4w(w) + 4w(4)
1152 = w[ 2w(18) + 2w(w) + 4(18) + 4(w) ]
=
1152 = w( 36w  2w2 + 72  4w )
+ 16w
4w2
1152 = w( 2w2 + 32w + 72 )
1152 = w( 2w2 ) + w(32w) + w(72)
A polynomial that represents the volume of the birdhouse
is ( 4w2 + 16w ) cubic inches.
b.
length
1152 = (2w + 4)
128 = 4w2 + 16w  128
1152 = 2w3 + 32w2 + 72w
128  128 = 4w2 + 16w  128w
0=
4w2
1152  1152 = 2w3 + 32w2 + 72w  1152
+ 16w  128
0 = 2w3 + 32w2 + 72w  1152
0 = 4( w2 + 4w  32 )
1
—4 (0)
=
0=
1
—4
0 = 2( w3  16w2  36w + 576 )
[ 4( w2 + 4w  32 ) ]
—12 (0)
+ 4w  32
(
0 = w  4 )( w + 8 )
w2
w4=
+4
0
+4
or
= —12 (2)( w3  16w2  36w + 576 )
0 = ( w3  16w2  36w + 576 )
w+8=
8
w=4
0 = ( w3  16w2 ) + (36w + 576)
0
0 = w2(w  16)  36(w  16)
8
0 = (w  16)( w2  36 )
w = 8
Disregard w = 8 because a negative width does not
make sense, so 4 is the correct value of w.
height = w + 4
=4+4
=8
The length is 4 inches, the width is 4 inches, and the
height is 8 inches.
0 = (w  16)( w2  6 2 )
0 = (w  16)(w + 6)(w  6)
w  16 =
+ 16
0
w+6=
or
+ 16
6
w = 16
0
or w  6 =
6
+6
w = 6
0
+6
w=6
Disregard w = 6 because a negative width does not make
sense. You know the height is greater than the width. Test the
solutions of the equation, 16 and 6, in the expression for height.
height = 18  w = 18  16 = 2 ✗
or
height = 18  w = 18  6 = 12 ✓
The solution 6 gives a height of 12 inches, which is greater
than 6, so 6 is the correct value of w.
length = 2w + 4
= 2(6) + 4
= 12 + 4
= 16
The width is 6 inches, the length is 16 inches, and the height
is 12 inches.
37. x3 + 2x2y  x  2y = ( x3 + 2x2y ) + (x  2y)
= x2(x + 2y)  (x + 2y)
= (x + 2y)( x2  1 )
= (x + 2y)( x2  12 )
= (x + 2y)(x + 1)(x  1)
So,
x3
+
2x2y
 x  2y = (x + 2y)(x + 1)(x  1).
38. 8b3  4b2a  18b + 9a = ( 8b3  4b2a ) + (18b + 9a)
= 4b2(2b  a)  9(2b  a)
= (2b  a)( 4b2  9 )
= (2b  a)[ (2b) 2  3 2 ]
So,
442
Algebra 1
Worked-Out Solutions
8b3

4b2a
= (2b  a)(2b + 3)(2b  3)
 18b + 9a = (2b  a)(2b + 3)(2b  3).
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Chapter 7
39. 4s2  s + 12st  3t = ( 4s2  s ) + (12st  3t)
46. a. width = 4 + h
= s(4s  1) + 3t(4s  1)
length = 9  h
= (4s  1)(s + 3t)
So, 4s2  s + 2st  3t = (4s  1)(s + 3t).
Volume =
length
=
(9  h)
= ( m2  2n )(6m + n)
= h(36) + h(5h)  h( h2 )
41. no; x3 + 2x2 + 3x + 6 = 0
= 36h + 5h2  h3
( x3 + 2x2 ) + (3x + 6) = 0
= h3 + 5h2 + 36
x2(x + 2) + 3(x + 2) = 0
A polynomial that represents the volume of the box in
terms of its height is ( h3 + 5h2 + 36h ) cubic inches.
(x + 2)( x2 + 3 ) = 0
The factors of the polynomial are x + 2 and x2 + 3. Using
the zero product property, x + 2 = 0 will give one real
solution x = 2. However, x2 + 3 = 0 has no real solutions
because when you subtract 3 from each side you get
x2 = 3, and no real number multiplied by itself will give a
negative number.
180 +
h3

h3  5h2  36h + 180 = 0
(  5h2 ) + (36h + 180) = 0
h3
h2(h  5)  36(h  5) = 0
h5=
y = x(x  3)(x + 3)(x  2)
+5
0 = x(x  3)(x + 3)(x  2)
3
x=3
3
+2
0
or
h+6=
+5
6
h=5
0 or x  2 =
x = 3
 36h = h3 + h3 + 5h2  5h2 + 36h  36h
(h  5)(h + 6)(h  6) = 0
y = x4  2x3  9x2 + 18x
0 or x + 3 =
5h2
(h  5)( h2  62 ) = 0
equal to 0 and solve for x to get the x-coordinates of the
x-intercepts.
0
+2
x=2
The x-intercepts of the graph of the function are 0, 3, 3,
and 2.
43. a. Sample answer:
x2(x + 1) + 1(x + 2) = x2(x) + x2(1) + 1(x) + 1(2)
= x3 + x2 + x + 2
Because the original expression did not have a common
binomial, x3 + x2 + x + 2 is not factorable.
0
6
h = 6
or
h6=
+6
0
+6
h=6
Disregard h = 6 because a negative height does not
make sense. So, the height is either 5 inches or 6 inches.
If h = 5, the width = 4 + h = 4 + 5 = 9 and the
length = 9  h = 9  5 = 4.
If h = 6, the width = 4 + h = 4 + 6 = 10 and the
length = 9  h = 9  6 = 3.
So, one possibility is that the length is 4 inches, the width
is 9 inches, and the height is 5 inches. Another possibility
is that the length is 3 inches, the width is 10 inches, and
the height is 6 inches.
c. SA = 2ℓ w + 2ℓ h + 2wh
= 2(4)(9) + 2(4)(5) + 2(9)(5)
b. Sample answer:
x2(x
180 = h3 + 5h2 + 39h
b.
(h  5)( h2  36 ) = 0
42. The x-intercepts occur when y = 0, so set each factor
+3
h
= h( 36 + 5h  h2 )
So, 6m3  12mn + m2n  2n2 = ( m2  2n )(6m + n).
+3
height
= h[ 9(4) + 9(h)  h(4)  h(h) ]
= h( 36 + 9h  4h  h2 )
= 6m( m2  2n ) + n( m2  2n )
or x  3 =
⋅
⋅
width
(4 + h)
= h(9  h)(4 + h)
40. 6m3  12mn + m2n  2n2 = ( 6m3  12mn ) + ( m2n  2n2 )
x=0
⋅
⋅
+ 1) + 1(x + 1) =
x2(x)
+
x2(1)
+ 1(x) + 1(1)
= x3 + x2 + x + 1
Because the original expression had a common binomial,
x3 + x2 + x + 1 can be factored by grouping.
44. Your friend is not correct because it is possible that the terms
have a common monomial factor.
45. 12z3  27z = 3z( 4z2  9 )
= 3z[ (2z) 2  3 2 ]
= 3z(2z + 3)(2z  3)
Expressions that could represent the dimensions of the room
are 3z, 2z + 3, and 2z  3.
Copyright © Big Ideas Learning, LLC
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= 72 + 40 + 90
= 202
SA = 2ℓ w + 2ℓ h + 2wh
= 2(3)(10) + 2(3)(6) + 2(10)(6)
= 60 + 36 + 120
= 216
The box that has a length of 4 inches, a width of 9 inches,
and a height of 5 inches is the box with the least possible
surface area. These dimensions result in a surface area of
202 square inches. The other possible dimensions result in
a surface area of 216 square inches.
Algebra 1
Worked-Out Solutions
443
Chapter 7
V = π r 2h
47.
b. Sample answer: 5x3 + wx2 + 80x = 0
25hπ = π (h  3)2h
(
π
π
Factors of 16
25h = (h  3)2h
Sum of factors
25h = [ h2  2(h)(3) + 32 ]h
⋅
25h = h2(h)  6h(h) + 9h
25h = h3  6h2 + 9h
25h  25h = h3  6h2 + 9h  25h
0 = h( h2  6h  16 )
0 = h(h  8)(h + 2)
h8=
+8
2, 8
4, 4
17
10
8
⋅
The equation has three solutions if w = 50. When w = 50,
factoring out 5x will leave a factorable trinomial that
is not a perfect square trinomial, so there will be three
factors.
0 = h3  6h2  16h
or
1, 16
w
Let — = 10.
5
w
5 — = 5 10
5
w = 50
25h = ( h2  6h + 9 )h
h=0
)
w
5x x2 + — x + 16 = 0
5
25hπ = π (h  3)2h
0
or
+8
h=8
h+2=
2
0
2
h = 2
Disregard h = 0 and h = 2 because the height cannot be
zero or negative. So, the height is 8 units and the radius is
8  3 = 5 units.
48. x5  x 4  5x 3 + 5x 2 + 4x  4
= ( x5  x4 ) + ( 5x 3 + 5x 2 ) + (4x  4)
= x 4(x  1)  5x 2(x  1) + 4(x  1)
Maintaining Mathematical Prociency
50. Use m = 1 and b =4 to graph y = x  4.
Use m = 2 and b = 2 to graph y = 2x + 2.
y
y = 2x + 2
2
4
2
2
4 x
(2, 2)
2
y=x4
= (x  1)( x 4  5x 2 + 4 )
= (x  1)( x 2  4 )( x 2  1 )
The solution is the point of intersection, (2, 2).
= (x  1)( x 2  2 2 )( x 2  12 )
= (x  1)(x  2)(x + 2)(x  1)(x + 1)
So,
x5

x4

5x 3
+
5x 2
+ 4x  4
1
Use m = 3 and b = 3 to graph y = 3x  3.
= (x  1)2(x + 1)(x  2)(x + 2).
4
49. a. Sample answer: 5x 3 + wx 2 + 80x = 0
(
)
w
5x x 2 + — x + 16 = 0
5
w
Let — = 2(1)(4).
5
w
—=8
5
w
5 —=8 5
5
w = 40
⋅
1
51. Use m = —2 and b = 2 to graph y = —2 x + 2.
⋅
y
(2, 3)
y = 1x + 2
2
2
2
4
x
2
y = 3x  3
The solution is the point of intersection, (2, 3).
The equation has two solutions if w = 40. When w = 40,
factoring out 5x will leave a perfect square trinomial, so
there will be two factors.
444
Algebra 1
Worked-Out Solutions
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Chapter 7
()
1 x
1
5x  y = 12
—4 x + y = 9
5x  5x  y = 12  5x
—4 x  —4 x + y = 9  —4 x
52.
1
1
55. y = 9 —3
1
y = —14x + 9
y = 5x + 12
2
x
( ) 9( — )
1(y) = 1(5x + 12)
1 2
3
x
9 —13
Use m = 5 and b = 12 to graph y = 5x  12.
1
0
()
9 —13
27
9
9 —13
81
y
y = 5x  12
1
1
y
(4, 8)
1
8
4
4
8
4
8 x
4
The solution is the point of intersection, (4, 8).
y  10 = 2x
3y
3
x
3
—=—
1
3
y  10 + 10 = 2x + 10
—x = y
1 3
3
1 4
3
3
1
—3
1
—9
1
1 x
2
4
6 x
2
1
0
1
2
3(0.5)2
3(0.5)1
3(0.5)0
3(0.5)1
3(0.5)2
y
12
6
3
1.5
0.75
From the graph, you can see the
domain is all real numbers and
the range is y < 0.
y
y = 2x + 10
2
4
6 x
y = 3(0.5)x
8
12
y = 2x + 10
16
y = 1x
3
4
8
57. f(x) = 3(4)x
4 x
2
x
4
1
0
1
2
3(4)x 3(4)2 3(4)1 3(4)0 3(4)1 3(4)2
(6, 2)
The solution is the point of intersection, (6, 2).
f(x)
3
3
—
16
54. f (x) = 5x
—4
3
12
48
y
6
x
2
1
0
1
2
5x
52
51
50
51
52
f (x)
1
—
25
1
—5
1
5
25
16
From the graph, you can see the
domain is all real numbers and
the range is y > 0.
y
12
8
4
2
1 2
3
3(0.5)x
y
4
1 1
3
x
1
1
Use m = — and b = 0 to graph y = — x.
3
3
Use m = 2 and b = 10 to graph y = 2x + 10.
12
0
56. y = 3(0.5)x
y = 5x  12
53. x = 3y
4
y = 9( 3)
2
4
3
From the graph, you can see the
domain is all real numbers and
the range is y > 0.
Use m = —4 and b = 9 to graph y = —4 x + 9.
y = 4 x + 9
2
( ) 9( — ) 9( — ) 9( — ) 9( — )
y
1
1
4
2
2 x
8
f (x) = 3(4)x
16
24
32
From the graph, you can see the domain is all real numbers
and the range is y < 0.
7.5 –7.8 What Did You Learn? (p. 409)
f (x) = 5x
2
4 x
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All rights reserved.
1. The solutions of the equation m2 + 3m + 2 = 0 are the
m-coordinates of the m-intercepts of the graph of
y = m2 + 3m + 2.
Algebra 1
Worked-Out Solutions
445
Chapter 7
2. The solutions of the equation 280 = x2 + 38x are
x = 10 and x = 28. However, the diagram shows that the
width of the road is less than 18 meters. So, a width of
28 meters does not make sense in this situation. Therefore,
the only reasonable solution is x = 10. So, the width of the
road is 10 meters.
F
10.
= 3y 2  7y  40
11. (x + 4)( x2 + 7x ) = x( x2 ) + x(7x) + 4( x2 ) + 4(7x)
= x3 + 7x2 + 4x2 + 28x
= x3 + 11x2 + 28x
12. (3y + 1)( 4y 2  y  7 )
4y 2  y  7
The leading coefcient is 2.
×
The polynomial has 2 terms. So, it is a binomial.
y7
12y 3 + 3y 2 + 21y
form as 5p6  3p3  4.
The leading coefcient is 5.
3y + 1
4y 2
2. You can write the polynomial 3p3 + 5p6  4 in standard
The greatest degree is 6. So, the degree of the polynomial is 6.
12y 3 + 7y 2 + 20y  7
13. (x + 9)(x  9) = x2  92
= x2  81
The polynomial has 3 terms. So, it is a trinomial.
3. You can write the polynomial 9x7  6x2 + 13x5 in standard
form as
9x7
+
13x5

14. (2y + 4)(2y  4) = (2y)2  42
= 4y 2  16
6x2.
The greatest degree is 7. So, the degree of the polynomial is 7.
The leading coefcient is 9.
15. ( p + 4)2 = p 2 + 2( p)(4) + 42
= p 2 + 8p + 16
The polynomial has 3 terms. So, it is a trinomial.
4. You can write the polynomial 12y + 8y3 in standard form
L
= 3y 2 + 8y + (15y) + (40)
1. You can write the polynomial 6 + 2x2 in standard form as
The greatest degree is 2. So, the degree of the polynomial is 2.
I
(y  5)(3y + 8) = y(3y) + y(8) + (5)(3y) + (5)(8)
Chapter 7 Review (pp. 410–412)
2x2 + 6.
O
16. (1 + 2d )2 = (1)2 + 2(1)(2d) + (2d )2
= 1  4d + 4d 2
as 8y3  12y.
= 4d 2  4d + 1
The greatest degree is 3. So, the degree of the polynomial is 3.
17. x 2 + 5x = 0
The leading coefcient is 8.
x(x + 5) = 0
The polynomial has 2 terms. So, it is a binomial.
x=0
5. (3a + 7) + (a  1) = 3a + a + 7  1
or
x+5=0
5
= (3a + a) + (7  1)
= 4a + 6
6. (x2 + 6x  5) + (2x2 + 15) = x2 + 2x2 + 6x  5 + 15
= (x2 + 2x2) + 6x + (5 + 15)
The roots are x = 0 and x = 5.
18. (z + 3)(z  7) = 0
z+3=0
= 3x2 + 6x + 10
3
7. (y2 + y + 2)  (y2  5y  2)
= 2y2 + 6y + 4
or
3
z7=0
+7
z = 3
= y2 + y + 2  y2 + 5y + 2
= (y2  y2) + (y + 5y) + (2 + 2)
5
x = 5
+7
z=7
The roots are z = 3 and z = 7.
(b + 13)2 = 0
19.
(b + 13)(b + 13) = 0
8. ( p + 7)  (6p 2 + 13p) = p + 7  6p 2  13p
= 6p 2 + ( p  13p) + 7
= 6p 2  12p + 7
9. (x + 6)(x  4) = x(x  4) + 6(x  4)
b + 13 = 0
13
13
b = 13
or
b + 13 = 0
13
13
b = 13
The equation has repeated roots b = 13.
= x(x) + x(4) + 6(x) + 6(4)
= x2 + (4x) + 6x + (24)
= x2 + 2x  24
446
Algebra 1
Worked-Out Solutions
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Chapter 7
20. 2y( y  9)( y + 4) = 0
2y = 0
2y 0
—=—
2
2
y=0
or
26. 5y 2  22y  8 = (5y 2 + 22y + 8)
y9=0
y+4=0
or
+9 +9
4
y=9
Factors Factors
of 5
of 8
4
y = 4
The roots are y = 0, y = 9, and y = 4.
21. p 2 + 2p  35
Factors of −35
–1, 35
1, –35
–5, 7
5, –7
Sum of factors
34
–34
2
–2
So,
b2
1, 80
2, 40
4, 20
5, 16
8, 10
81
42
24
21
18
+ 18b + 80 = (b + 8)(b + 10).
Factors of −21
–1, 21
1, –21
–3, 7
3, –7
Sum of factors
20
–20
4
−4
So,
24.
x2
1, 8
(y + 1)(5y + 8)
8y + 5y = 13y
✗
1, 5
8, 1
(y + 8)(5y + 1)
y + 40y = 41y
✗
1, 5
2, 4
(y + 2)(5y + 4)
4y + 10y = 14y
✗
1, 5
4, 2
(y + 4)(5y + 2)
2y + 20y = 22y
✓
Factors Factors
of 6
of 7
23. z 2  4z  21
z2
1, 5
27. 6x 2 + 17x + 7
22. b 2 + 18b + 80
Sum of factors
–1, –28
–2, –14
–4, –7
–29
–16
–11
25. 3t 2 + 16t  12
Possible
factorization
1, 6
1, 7
(x + 1)(6x + 7)
7x + 6x = 13x
✗
1, 6
7, 1
(x + 7)(6x + 1)
x + 42x = 43x
✗
2, 3
1, 7
(2x + 1)(3x + 7)
14x + 3x = 17x
✓
2, 3
7, 1
(2x + 7)(3x + 1)
2x + 21x = 23x
✗
Factors Factors
of 2
of 6
So, x2  11x + 28 = (x  4)(x  7).
Factors Factors
of 3
of −12
Middle term
28. 2y2 + 7y  6 = (2y2  7y + 6)
 11x + 28
Sum of factors
Possible
factorization
So, 6x 2 + 17x + 7 = (2x + 1)(3x + 7).
 4z  21 = (z + 3)(z  7).
Factors of 28
Middle term
So, 5y2  22y  8 = (y + 4)(5y + 2).
So, p 2 + 2p  35 = (p  5)(p + 7).
Factors of 80
Possible
factorization
Possible
factorization
Middle term
1, 2
–1, –6
(y – 1)(2y – 6)
–6y – 2y = –8y
✗
1, 2
–6, –1
(y – 6)(2y – 1)
–y – 12y = –13y
✗
1, 2
–2, –3
(y – 2)(2y – 3)
–3y – 4y = –7y
✓
1, 2
–3, –2
(y – 3)(2y – 2)
–2y – 6y = –8y
✗
So, 2y 2 + 7y  6 = (y  2)(2y  3).
29. 3z 2 + 26z  9
Middle term
1, 3
1, –12
(t + 1)(3t – 12)
–12t + 3t = –9t
✗
1, 3
12, –1
(t + 12)(3t – 1)
–t + 36t = 35t
✗
1, 3
–1, 12
(t – 1)(3t + 12)
12t – 3t = 9t
✗
1, 3
1, 3
–12, 1
(t – 12)(3t + 1)
✗
1, 3
2, –6
(t + 2)(3t – 6)
–6t + 6t = 0
1, 3
6, –2
(t + 6)(3t – 2)
1, 3
–2, 6
(t – 2)(3t + 6)
1, 3
–6, 2
(t – 6)(3t + 2)
1, 3
3, –4
(t + 3)(3t – 4)
–4t + 9t = 5t
✗
1, 3
4, –3
(t + 4)(3t – 3)
–3t + 12t = 9t
✗
1, 3
–3, 4
(t – 3)(3t + 4)
4t – 9t = –5t
✗
1, 3
–4, 3
(t – 4)(3t + 3)
9t – 12t = –3t
✗
Possible
factorization
Middle term
1, –9
(z + 1)(3z – 9)
–9z + 3z = –6z
✗
1, 3
9, –1
(z + 9)(3z – 1)
–z + 27z = 26z
✓
✗
1, 3
–1, 9
(z – 1)(3z + 9)
9z – 3z = 6z
✗
–2t + 18t = 16t
✓
1, 3
–9, 1
(z – 9)(3z + 1)
z – 27z = –26z
✗
6t – 6t = 0
✗
1, 3
3, –3
(z + 3)(3z – 3)
✗
1, 3
–3, 3
(z – 3)(3z + 3)
t – 36t = –35t
2t – 18t = –16t
Factors Factors
of 3
of −9
–3z + 9z = 6z
3z – 9z = –6z
✗
✗
So, 3z 2 + 26z  9 = (z + 9)(3z  1).
So, 3t 2 + 16t  12 = (t + 6)(3t  2).
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Algebra 1
Worked-Out Solutions
447
Chapter 7
30. 10a 2  13a  3
38. 3x 3  9x 2  54x = 0
Factors Factors
of 10
of −3
3x(x 2  3x  18) = 0
Possible
factorization
3x(x  6)(x + 3) = 0
Middle term
1, 10
1, –3
(a + 1)(10a – 3)
–3a + 10a = 7a ✗
1, 10
3, –1
(a + 3)(10a – 1)
–a + 30a = 29a ✗
1, 10
–1, 3
(a – 1)(10a + 3)
3a – 10a = –7a
✗
3x = 0
3x 0
—=—
3
3
x=0
1, 10
–3, 1
(a – 3)(10a + 1)
a – 30a = –29a
✗
The roots are x = 0, x = 6, and x = 3.
2, 5
1, –3
(2a + 1)(5a – 3)
–6a + 5a = –a
✗
2, 5
3, –1
(2a + 3)(5a – 1) –2a + 15a = 13a ✗
2, 5
–1, 3
(2a – 1)(5a + 3)
2, 5
–3, 1
(2a – 3)(5a + 1)
2a – 15a = –13a ✓
3
3
x = 3
or
2x  3 = 0
3
+3
+3
2x = 3
2x 3
—=—
2
2
3
x = —
2
2x = 3
2x 3
—=—
2
2
3
x=—
2
3
3
The roots are x = — and x = —.
2
2
y 2  100 = y 2  102
= (y + 10)(y  10)
So, y 2  100 = (y + 10)(y  10).
z 2  6z + 9 = z2  2(z)(3) + 32
= (z  3)2
40.
So, z 2  6z + 9 = (z  3)2.
z 3 + 3z 2  25z  75 = 0
(z 3 + 3z 2) + (25z  75) = 0
z 2(z + 3)  25(z + 3) = 0
m 2 + 16m + 64 = m 2 + 2(m)(8) + 82
(z + 3)(z 2  25) = 0
= (m + 8)2
(z + 3)(z 2  52) = 0
So, m 2 + 6m + 64 = (m + 8)2.
(z + 3)(z + 5)(z  5) = 0
n 3  9n = n(n 2  9)
35.
x=6
2x + 3 = 0
So, x 2  9 = (x + 3)(x  3).
34.
3
(2x + 3)(2x  3) = 0
= (x + 3)(x  3)
33.
+6
(2x)2  32 = 0
x 2  9 = x 2  32
32.
+6
x+3=0
or
4(4x 2  9) = 0
1
1
— [ 4(4x 2  9) ] = —(0)
4
4
4x 2  9 = 0
So, 10a 2  13a  3 = (2a  3)(5a + 1).
31.
x6=0
16x 2  36 = 0
39.
✗
6a – 5a = a
or
z+3=0
= n(n 2  32)
3
= n(n + 3)(n  3)
3
z = 3
So, n 3  9n = n(n + 3)(n  3).
or
z+5=0
5
or
5
z = 5
z5=0
+5 +5
z=5
The roots are z = 3, z = 5, and z = 5.
x 2  3x + 4ax  12a = (x 2  3x) + (4ax  12a)
36.
= x(x  3) + 4a(x  3)
= (x  3)(x + 4a)
So, x 2  3x + 4ax  12a = (x  3)(x + 4a).
37. 2x 4 + 2x 3  20x 2 = 2x 2(x 2 + x  10)
Factors of −10
–1, 10
1, –10
–2, 5
2, –5
Sum of factors
9
–9
3
–3
+ x  10 is unfactorable, so the original polynomial is
factored completely. So, 2x4 + 2x3  20x2 = 2x2(x2 + x  10).
x2
448
Algebra 1
Worked-Out Solutions
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Chapter 7
⋅ width ⋅ height
96 = (x + 8) ⋅ x ⋅ (x  2)
F
5.
41. Volume = length
= 6w 2 + w  15
So, (2w  3)(3w + 5) = 6w 2 + w  15.
96 = x[ x 2 + (2x + 8x)  16 ]
96 = x(x 2 + 6x  16)
+ x(6x) + x(16)
96  96 =
+
6x 2
= z 2  121
So, (z + 11)(z  11) = z 2  121.
 16x  96
7. Check the rst and last terms to verify they are perfect
0 = (x 3 + 6x 2) + (16x  96)
squares. If they are, nd 2 times the product of their square
roots. If this matches the middle term, or the opposite of the
middle term, it is a perfect square trinomial.
0 = x 2(x + 6)  16(x + 6)
0 = (x + 6)(x 2  16)
0 = (x + 6)(x 2  42)
8. A monomial is a type of polynomial, and a number is a type
0 = (x + 6)(x + 4)(x  4)
x+6=0
6
x+4=0
or
6
4
x = 6
(z + 11)(z  11) = z 2  112
6.
96 = x 3 + 6x 2  16x
x3
of monomial. So, 18 is a polynomial.
or
x4=0
4
+4
x = 4
+4
9. s 2  15s + 50
Factors of 50
x=4
Disregard x = 6 and x = 4 because a negative width
does not make sense. So, the solution is x = 4, which means
the length of the box is 4 + 8 = 12 feet, the width is 4 feet,
and the height is 4  2 = 2 feet.
Sum of factors
–1, –50 –2, –25 –5, –10
–51
–27
–15
So, s 2  15s + 50 = (s  5)(s  10).
h 3 + 2h 2  9h  18 = (h 3 + 2h 2) + (9h  18)
10.
= h 2(h + 2)  9(h + 2)
Chapter 7 Test (p. 413)
= (h + 2)(h 2  9)
1. (2p + 4)  (p 2  6p + 8) = 2p + 4  p 2 + 6p  8
= (h + 2)(h 2  32)
= p2 + (2p + 6p) + (4  8)
= (h + 2)(h + 3)(h  3)
= p 2 + 4p  4
The greatest degree is 2. So, the degree of the polynomial is 2.
The polynomial has 3 terms. So, it is a trinomial.
2. (9c6  5b 4)  (4c6  5b 4) = 9c6  5b 4  4c6 + 5b 4
= (9c6  4c6) + (5b 4 + 5b 4)
So, h 3 + 2h 2  9h  18 = (h + 2)(h + 3)(h  3).
11. 5k 2  22k + 15 = (5k 2 + 22k  15)
Factors Factors
of 5
of −15
Possible
factorization
Middle term
= 5c6 + 0
1, 5
1, –15
(k + 1)(5k – 15) –15k + 15k = 10k
✗
=
1, 5
15, –1
(k + 15)(5k – 1)
–k + 75k = 74k
✗
1, 5
–1, 15
(k – 1)(5k + 15)
15k – 5k = 10k
✗
1, 5
–15, 1
(k – 15)(5k + 1)
1, 5
3, –5
(k + 3)(5k – 5)
–5k + 15k = 10k
✗
1, 5
5, –3
(k + 5)(5k – 3)
–3k + 25k = 22k
✓
= 6s 4  3t
1, 5
–3, 5
(k – 3)(5k + 5)
5k – 15k = –10k ✗
The greatest degree is 4. So, the degree of the polynomial is 4.
1, 5
–5, 3
(k – 5)(5k + 3)
3k – 25k = –22k ✗
The polynomial has 2 terms. So, it is a binomial.
So, 5k 2  22k + 15 = (k + 5)(5k  3).
5c6
The expression has a degree of 6.
The polynomial has 1 term. So, it is a monomial.
3. (4s 4 + 2st + t) + (2s 4  2st  4t)
= (4s 4 + 2s 4) + (2st  2st) + (t  4t)
= 6s 4 + 0 + (3t)
4.
L
= 6w 2 + (10w  9w) + (15)
96 = x[x(x) + x(2) + 8(x) + 8(2)]
96 =
I
(2w  3)(3w + 5) = 2w(3w) + 2w(5) + (3)(3w) + (3)(5)
96 = x(x + 8)(x  2)
x(x 2)
O
k – 75k = –74k ✗
(h  5)(h  8) = h(h  8)  5(h  8)
= h(h) + h(8) + (5)(h) + (5)(8)
= h2 + (8h  5h) + 40
= h2  13h + 40
So, (h  5)(h  8) = h2  13h + 40.
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Algebra 1
Worked-Out Solutions
449
Chapter 7
12. (n  1)(n + 6)(n + 5) = 0
n1=0
+1
16. a. x 2 + 27x + 176
n+6=0
or
+1
6
n=1
n+5=0
or
6
5
n = 6
5
n = 5
d 2 + 2(d)(7) + 72 = 0
c.
7
d = 7
+
=
11, 16
Sum of factors
177
90
48
30
27
⋅
Area = length width
1
10 = 2x x + —
2
1
10 = 2x x + —
2
1
10 = 2x(x) + 2x —
2
10 = 2x 2 + x
⋅(
(
The equation has repeated roots of d = 7.
14.
8, 22
2x + 32 + 2x + 22 = (4x + 54) feet.
d+7=0
8x 2
4, 44
b. The perimeter is P = 2(x + 16) + 2(x + 11) =
(d + 7)2 = 0
6x 4
2, 88
Because the length of the stage is represented by the
binomial (x + 16) feet, a binomial for the width of the
stage is (x + 11) feet.
d 2 + 14d + 49 = 0
7
1, 176
So, x 2 + 27x + 176 = (x + 11)(x + 16).
The roots are n = 1, n = 6, and n = 5.
13.
Factors of 176
26x 3
6x 4 + 8x 2  26x 3 = 26x 3  26x 3
)
)
()
6x 4  26x 3 + 8x 2 = 0
10  10 = 2x 2 + x  10
2x2(3x 2  13x + 4) = 0
0 = 2x 2 + x  10
Factors Factors
of 3
of 4
–1, –4
1, 3
Possible
factorization
(x – 1)(3x – 4)
Factors Factors
Possible
of 2
of −10 factorization
Middle term
–4x – 3x = –7x
✗
1, 3
–4, –1
(x – 4)(3x – 1)
–x – 12x = –13x
✓
1, 3
–2, –2
(x – 2)(3x – 2)
–2x – 6x = –8x
✗
2x 2(x  4)(3x  1) = 0
2x 2 = 0
2x 2 0
—=—
2
2
x2 = 0
—
or
x4=0
+4
or
+4
x=4
—
√x2 = √0
x=0
3x  1 = 0
+1
1, 2
1, –10
(x + 1)
(2x – 10)
–10x + 2x = –8x
1, 2
10, –1
(x + 10)
(2x – 1)
–x + 20x = 19x
1, 2
–1, 10
(x – 1)
(2x + 10)
10x – 2x = 8x
1, 2
–10, 1
(x – 10)
(2x + 1)
1, 2
2, –5
(x + 2)
(2x – 5)
–5x + 4x = –x
1, 2
5, –2
(x + 5)
(2x – 2)
–2x + 10x = 8x
1, 2
–2, 5
(x – 2)
(2x + 5)
5x – 4x = x
1, 2
–5, 2
(x – 5)
(2x + 2)
2x – 10x = –8x
+1
3x = 1
3x 1
—=—
3
3
1
x=—
3
1
The equation has roots of x = 4, x = —, and repeated roots
3
of x = 0.
15. π(r  3)2 = π(r  3)(r  3)
= π[ r(r) + r(3) + (3)(r) + (3)(3) ]
= π[ r 2 + (3r  3r) + 9 ]
= π (r 2  6r + 9)
= π (r 2)  π (6r) + π (9)
= πr 2  6πr + 9π
So, a polynomial that represents the area covered by the
hour hand of the clock in one rotation is (πr 2  6πr + 9π)
square units.
Middle term
✗
✗
✗
✗
x – 20x = –19x
✗
✗
✓
✗
0 = (x  2)(2x + 5)
x–2=0
+2
+2
or
2x + 5 = 0
–5
–5
x=2
2x = –5
–5
2x
— =—
2
2
5
x=–—
2
5
5
Disregard x = — because 2x = 2 — = 5 feet does
2
2
not make sense as the length of the trap door. So, the
solution is x = 2.
( )
450
Algebra 1
Worked-Out Solutions
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Chapter 7
d. Area of stage = x 2 + 27x + 176
= (2)2 + 27(2) + 176
= 4 + 54 + 176
= 234 ft2
(
1
Area of trap door = 2x x + —
2
)
( )
()
1
= 2(2) 2 + —
2
5
= 2(2) —
2
= 10 ft2
⋅
20 Area of trapdoor = 20(10)
= 200
ft2
The area of the stage is 234 square feet, and 20 times the
area of the trap door is 200 square feet. So, the magician’s
requirement is satised because the area of the stage is
greater than 20 times the area of the trap door.
17. Sample answer: A polynomial equation in factored form that
has three positive roots is (x  1)(x  2)(x  3) = 0.
18. Let y = 0.
x 3 + 3x 2  16x  48 = 0
( x 3 + 3x 2 ) + (16x  48) = 0
x 2(x + 3)  16(x + 3) = 0
(x + 3)( x 2  16 ) = 0
(x + 3)( x 2  42 ) = 0
(x + 3)(x + 4)(x  4) = 0
x+3=0
3
3
x = 3
or
x+4=0
4
or
x4=0
4
+4
x = 4
+4
x=4
Disregard x = 3 and x = 4 because widths of
3  2 = 5 and 4  2 = 6, respectively, do not make
sense. So, the solution is x = 4.
length = x + 6
=4+6
= 10
width = x  2
=2
0 = 16t 2 + 24t
height = x  1
0 = 8t(2t  3)
=41
2t  3 = 0
or
x 3 + 3x 2  16x + 12  60 = 60  60
=42
y = 16t 2 + 24t
8t = 0
0
8t
—=—
8
8
t=0
x 3 + 3x2  16x + 12 = 60
b.
+3
+3
2t = 3
2t 3
—=—
2
2
t = 1.5
You leave the surface of the trampoline when t = 0, then you
are in the air, and then you make contact with the surface of
the trampoline again after 1.5 seconds. So, you are in the air
for 1.5  0 = 1.5 seconds.
=3
So, the length of the box is 10 inches, the width is 2 inches,
and the height is 3 inches.
Chapter 7 Standards Assessment (pp. 414–415)
1. a. The polynomial 4x 3 has 1 term. So, it is a monomial.
b. The polynomial 6y  3y5 has 2 terms. So, it is a
binomial.
c. The polynomial c2 + 2 + c has 3 terms. So, it is a
trinomial.
⋅ width ⋅ height
= (x + 6) ⋅ (x  2) ⋅ (x  1)
d. The polynomial 10d 4 + 7d 2 has 2 terms. So, it is a
= (x + 6)[x(x) + x(1) + (2)(x) + (2)(1)]
f. The polynomial 3b6  12b8 + 4b4 has 3 terms. So, it is a
19. a. Volume = length
= (x + 6)(x  2)(x  1)
= (x + 6)[ x 2 + (x  2x) + 2 ]
= (x + 6)( x 2  3x + 2 )
= x(x2) + x(3x) + x(2) + 6( x2 ) + 6(3x) + 6(2)
= x 3  3x 2 + 2x + 6x 2  18x + 12
= x 3 + ( 3x 2 + 6x2 ) + ( 2x  18x ) + 12
=
x3
+
3x 2
 16x + 12
So, a polynomial that represents the volume of the box is
( x 3 + 3x 2  16x + 12 ) cubic inches.
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binomial.
e. The polynomial 5z11 + 8z12 has 2 terms. So, it is a
binomial.
trinomial.
The order of the polynomials by degree from least to greatest
is c, a, d, b, f, and e.
2. D; Over the interval x = 0 to x = 2, the function f increases
by 25  4 = 21, the function g increases by 18  8 = 10,
the function h increases by 9  1 = 8, and the function j
increases by 72  2 = 70. So, the function j increases at the
fastest rate over this interval.
Algebra 1
Worked-Out Solutions
451
Chapter 7
x 3 + 6x 2  4x = 24
3.
6. 2x + 15x 2  8 = 15x 2  2x  8
x 3 + 6x 2  4x  24 = 24  24
Factors Factors
of 15
of −8
x 3 + 6x 2  4x  24 = 0
( x3 + 6x2 ) + (4x  24) = 0
x 2(x + 6)  4(x + 6) = 0
(x + 6)( x 2  4 ) = 0
(x + 6)( x 2  22 ) = 0
(x + 6)(x + 2)(x  2) = 0
x+6=0
6
or
x+2=0
6
2
x = 6
x2=0
or
2
+2
x = 2
+2
x=2
The solutions of the equation are x = 6, x = 2, and
x = 2.
+1
4.
+1
+1
+1
+1
Hours, x
1
2
3
4
5
6
Distance (miles), y
62
123
184
245
306
367
+61 +61 +61 +61 +61
As x increases by 1, y increases by 61. So there is a constant
rate of change, which means the table can be represented by
61
a linear function with m = —
= 61.
1
y  y1 = m( x  x1 )
y  62 = 61(x  1)
y  62 = 61(x)  61(1)
Possible
factorization
1, 15
1, 8
(x + 1)(15x  8)
8x + 15x = 7x
✗
1, 15
8, 1
(x + 8)(15x  1)
x + 120x = 119x
✗
1, 15
1, 8
(x  1)(15x + 8)
8x  15x = 7x
✗
1, 15
8, 1
(x  8)(15x + 1)
x  120x = 119x ✗
1, 15
2, 4
(x + 2)(15x  4)
4x + 30x = 26x
✗
1, 15
4, 2
(x + 4)(15x  2)
2x + 60x = 58x
✗
1, 15
2, 4
(x  2)(15x + 4)
4x  30x = 26x
✗
1, 15
4, 2
(x  4)(15x + 2)
2x  60x = 58x
✗
3, 5
1, 8
(3x + 1)(5x  8)
24x + 5x = 19x
✗
3, 5
8, 1
(3x + 8)(5x  1)
3x + 40x = 37x
✗
3, 5
8, 1
(3x  1)(5x + 8)
24x + 5x = 19x
✗
3, 5
8, 1
(3x  8)(5x + 1)
3x  40x = 37x
✗
3, 5
2, 4
(3x + 2)(5x  4) 12x + 10x = 2x
✓
3, 5
4, 2
(3x + 4)(5x  2)
6x + 20x = 14x
✗
3, 5
2, 4
(3x + 2)(5x + 4)
12x  10x = 2x
✗
3, 5
4, 2
(3x  4)(5x + 2)
7. a. The graph is going up from left to right. So, the function
y = 61x + 1
is increasing for increasing values of x.
So, an equation that models the distance traveled y as a
function of the number of hours x is y = 61x + 1.
5. a. Use m =
4
and b = 2 to graph y =
1
—3 x
b. From the graph, you can see that the range of the function
is y > 0. So, the graph does not have an x-intercept. The
graph appears to cross the y-axis at (0, 2). To verify,
check that f(0) equals 2: f(0) = 2(3)0 = 2(1) = 2. So, the
y-intercept is 2.
+ 2.
y
8. B;
x 2 + 6x  2
×
4
2
✗
= (3x + 2)(5x  4) = (5x  4)(3x + 2).
+ 62
1
—3
6x  20x = 18x
So, 2x + 15x 2  8 = 15x2  2x  8
y  62 = 61x  61
+62
Middle term
2
4 x
2x  4
4x 2  24x + 8
2
2x3 +12x 2  4x
4
2x 3 + 8x 2  28x + 8
b. The equation is of the form y = mx + b, and the graph is
a line. So, the equation represents a linear function.
c. There are no breaks in the graph. So, it is continuous.
452
Algebra 1
Worked-Out Solutions
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Chapter 7
9. a.
Area of
golf hole
Area of
Area of
= square tee area + rectangular area
= x 2 + 3x(x + 4)
= x 2 + 3x(x) + 3x(4)
= x 2 + 3x 2 + 12x
= 4x 2 + 12x
A polynomial that represents the area of the golf hole is
( 4x 2 + 12x ) square feet.
b. Perimeter
= x + x + (3x  x) + (x + 4) + 3x + [ (x + 4) + x ]
= x + x + 2x + x + 4 + 3x + 2x + 4
= x + x + 2x + x + 3x + 2x + 4 + 4
= 10x + 8
A polynomial that represents the perimeter of the golf
hole is (10x + 8) feet.
4x 2 + 12x = 216
c.
4x 2
+ 12x  216 = 216  216
4x 2 + 12x  216 = 0
4( x 2 + 3x  54 ) = 0
1
1
—[ 4( x 2 + 3x  54 ) ] = —(0)
4
4
x 2 + 3x  54 = 0
Factors
of −54
1,
2,
6,
1,
2,
3,
3,
9
54 54 27 27 18 18
6,
9
Sum of
factors
53
25
3
53
25
15
15
3
(x  6)(x + 9) = 0
x6=0
or
+6 +6
x=6
x+9=0
9
9
x = 9
Disregard x = 9 because a negative side length does not
make sense. So, the solution is x = 6.
Perimeter = 10x + 8
= 10(6) + 8
= 60 + 8
= 68
So, the perimeter of the golf hole is 68 feet.
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Algebra 1
Worked-Out Solutions
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