Review Sheet for Test 1
Math 101
9–11–2008
These problems are provided to help you study. The presence of a problem on this handout does not
imply that there will be a similar problem on the test. And the absence of a topic does not imply that it
won’t appear on the test.
1. (a) Solve for x: 2x + 1 = 4x − 5.
(b) Solve for x: 3(x − 2) = 5(x + 1).
(c) Solve for x:
1
1
1
x − (x − 1) = x + 5.
2
3
6
(d) Solve for x: 0.35(x + 4) = 0.5(0.7x + 2.8).
2. The perimeter of a rectangle is 50 meters. The length is 1 meter more than twice the width. Find the
dimensions.
3. The largest angle in a triangle is 4 degrees less than 3 times the smallest angle. The third angle is 29
degrees greater than the smallest angle. Find the angles.
4. (a) If Calvin has 6% more money than Bonzo, and Bonzo has $24.50, how much money does Calvin have?
(b) If Calvin has 6% more money than Bonzo, and Calvin has $14.31, how much money does Bonzo have?
(c) Which costs more all together: A $150 TV set with 6.5% sales tax or a $152 TV set with 6% sales tax?
(d) 3% of what is equal to 4.5% of 60?
5. (a) (Ideal gas law) Solve for T in pV = nRT .
(b) (Area of a trapezoid) Solve for b2 in A =
1
h(b1 + b2 ).
2
(c) Solve for c in abc + 2bc − 3abP = 0.
(d) Solve for k in ak = b(k + 1).
6. (a) Solve 3x + 5 > 4(2x − 1). Write your answer using inequality notation, and draw a picture of the
solution set.
(b) Solve −3x − 4 ≤ −8(x + 5). Write your answer using inequality notation, and draw a picture of the
solution set.
(c) Solve 0.5(6x − 4) < 3x − 10. Write your answer using inequality notation, and draw a picture of the
solution set.
(d) Solve −4 < 2(x + 1) < 10. Write your answer using inequality notation, and draw a picture of the
solution set.
7. (a) Solve for x: |x − 10| = 5.
(b) Solve for x: 5 + |x + 7| = 13.
(c) Solve for x: |x − 4| = −6.
(d) Solve for x: |2x + 3| = 17.
8. (a) Solve for x: |x − 1| < 4.
1
(b) Solve for x: |x + 6| ≥ 3.
(c) Solve for x: |x − 17| < −2.
(d) Solve for x: |2x − 5| > 13.
(e) Solve for x: | − x − 6| ≤ 1.
9. Calvin Butterball has $4.00 in dimes and nickels. The number of dimes is 5 less than twice the number of
nickels. Find the number of dimes and the number of nickels.
10. Silas Hogwinder has some 13-cent stamps, some 17-cent stamps, and some 40-cent stamps. The number
of 13-cent stamps is 2 less than the number of 17-cent stamps. The number of 40-cent stamps is 1 more than
twice the number of 17-cent stamps. The total value of the stamps is $13.34. Find the number of each type
of stamp that Silas has.
11. How many gallons of a 40% alcohol solution and a 30% alcohol solution must be mixed together to make
50 gallons of a 32% alcohol solution?
12. How many pounds of dried fruit worth $7 per pound must be mixed with 4 pounds of peppermint ketchup
worth $3.50 per pound to make a mixture worth $5 per pound?
13. Calvin drives down a long straight road in his 1978 Chevette at a constant speed of 30 miles an hour.
Phoebe starts out 50 miles behind him, and follows him at 70 miles an hour in her Porsche. How long will
it take before Phoebe runs Calvin off the road?
14. Leopold drives 6 miles per hour faster than Molly. If they start at the same point and drive in opposite
directions for 4 hours, they will be 272 miles apart. What is Molly’s speed?
15. Phoebe invests $2000 in two accounts. One pays 6% simple interest, while the other pays 8% simple
interest. At the end of a year, she has earned $128 in interest. How much was invested in each account?
16. The largest angle in a triangle is 16◦ less than three times the smallest angle. The third angle is 16◦
more than the smallest angle. Find the measures of the angles.
17. Multiply the polynomials:
(a) (2x − 3)(x + 4).
(b) (3x2 − x − 1)(x − 2).
(c) (2x − 5)(2x + 5).
(d) (3w + 7)2 .
(e) (1 − 6t)2 .
(f) (2x + y)(x − 3y).
(g) (x2 − y 4 )2 .
(h) (x + y + 1)(x + 2y).
18. Factor completely:
(a) x2 − 64.
(b) x2 + 8x + 16.
(c) 4x2 − 25.
(d) x2 − 5x + 6.
2
(e) x2 − 9x − 10.
(f) 4x2 + 8x − 5.
(g) a4 − 16b4 .
(h) x3 − 8y 3 .
(i) x3 + 5x2 − 16x − 80.
(j) x2 y − 3xy + xy 2 − 3y 2 .
(k)
81
− 1.
a2 b2
(l)
4x2
4x
+
+ 1.
y2
y
(m)a2 − 3a − ab + 3b.
(n) x3 + x2 − xy 2 − y 2 .
(o) x2 + xy − 5x − 5y.
(p) 2x3 + 7x2 + 3x.
19. Factor completely.
(a) x3 + 27y 3 .
(b)
1 3
a − 125b3 .
8
(c) 64 − y 3 .
(d) ab + 6b + a2 + 6a.
(e) x3 − 3x2 − 4x + 12.
(f) x2 y − 2xy 2 + x − 2y.
20. (a) Divide x4 + x3 + 1 by x2 − 1 using long division.
(b) Compute the quotient and remainder when 2x4 + 3x2 + 4x is divided by 2x2 .
(c) Compute the quotient and remainder when 2x2 + 5xy − 3y 2 is divided by x + 3y.
(d) Factor x3 − 2x2 − 11x + 12 completely, given that x − 4 is one of the factors.
Solutions to the Review Sheet for Test 1
1. (a) Solve for x: 2x + 1 = 4x − 5.
3
2x + 1
5
2x + 6
− 2x
6
÷
2
3
= 4x − 5
5
= 4x
2x
= 2x
2
= x
+
The solution is x = 3.
(b) Solve for x: 3(x − 2) = 5(x + 1).
3x
3(x − 2)
−
6
5
11
−
3x
− 3x
−
−11
2
11
−
2
÷
The solution is x = −
(c) Solve for x:
=
= 5x
5(x + 1)
+
5
5
= 5x
3x
= 2x
2
=
x
11
.
2
1
1
1
x − (x − 1) = x + 5.
2
3
6
−
1
x −
2
1
x −
2
1
x
6
1
x
6
1
x
3
1
(x − 1)
3
+
+
=
1
3
1
3
1
3
The last equation “
=
=
=
1
x +
6
1
x +
6
1
x +
6
1
x
6
5
5
5
5
1
= 5” is a contradiction. Therefore, the original equation has no solutions.
3
(d) Solve for x: 0.35(x + 4) = 0.5(0.7x + 2.8).
0.35x
− 0.35x
0.35(x + 4)
+
=
1.4 = 0.35x
1.4
0.35x
0 =
0
0.5(0.7x + 2.8)
+
1.4
1.4
The last equation is an identity. Therefore, the original equation is true for all x.
2. The perimeter of a rectangle is 50 meters. The length is 1 meter more than twice the width. Find the
dimensions.
Let L be the length and let W be the width.
The perimeter is 50 meters: 50 = 2L + 2W .
4
The length is 1 meter more than twice the width: L = 1 + 2W .
Plug L = 1 + 2W into 50 = 2L + 2W and simplify:
50 = 2(1 + 2W ) + 2W,
Solve for W :
50 = 2 + 4W + 2W,
50 =
2
48 =
÷ 6
8 =
50 = 2 + 6W.
2 + 6W
2
6W
6
W
−
W = 8 gives L = 1 + 2W = 1 + 2 · 8 = 17. The width is 8 meters and the length is 17 meters.
3. The largest angle in a triangle is 4 degrees less than 3 times the smallest angle. The third angle is 29
degrees greater than the smallest angle. Find the angles.
Let L be the largest angle, let S be the smallest angle, and let T be the third angle.
The sum of the angles in a triangle is 180 degrees: 180 = L + S + T .
The largest angle is 4 degrees less than 3 times the smallest angle: L = 3S − 4.
The third angle is 29 degrees greater than the smallest angle: T = 29 + S.
Substitute T = 29 + S and L = 3S − 4 into 180 = L + S + T and simplify:
180 = (3S − 4) + S + (29 + S),
Solve for S:
180 = 5S
25
155 = 5S
5
5
31 = S
−
÷
+
180 = 5S + 25.
25
25
S = 31 gives L = 3S − 4 = 3 · 31 − 4 = 89 and T = 29 + S = 29 + 31 = 60. The smallest angle is 31◦ ,
the largest angle is 89◦ , and the third angle is 60◦ .
4. (a) If Calvin has 6% more money than Bonzo, and Bonzo has $24.50, how much money does Calvin have?
Calvin has 6% more money than Bonzo:
(Calvin) = (Bonzo) + 6% of
(Bonzo),
(Bonzo) + 0.06 · (Bonzo),
(Calvin) = 1.06(Bonzo).
Since (Bonzo) = 24.50,
(Calvin) = 1.06 · 24.50 = 25.97.
Calvin has $25.97.
(b) If Calvin has 6% more money than Bonzo, and Calvin has $14.31, how much money does Bonzo have?
Calvin has 6% more money than Bonzo:
(Calvin) = (Bonzo) + 6% of
(Bonzo),
(Bonzo) + 0.06 · (Bonzo),
Since (Calvin) = 14.31,
14.31 = 1.06(Bonzo).
5
(Calvin) = 1.06(Bonzo).
Solve for (Bonzo):
÷
14.31 = 1.06(Bonzo)
1.06
1.06
13.50 =
(Bonzo)
Bonzo has $13.50.
(c) Which costs more all together: A $150 TV set with 6.5% sales tax or a $152 TV set with 6% sales tax?
A $150 TV set with 6.5% sales tax costs
150 + 6.5% of
150 = 150 + 0.065 · 150 = 159.75
A $152 TV set with 6% sales tax costs
152 + 6% of
152 = 152 + 0.06 · 152 = 161.12.
The $152 TV set with 6% sales tax costs more all together.
(d) 3% of what is equal to 4.5% of 60?
3% of what is equal to 4.5% of 60:
3% of
(what) = 4.5%
of 60,
0.03 · (what) = 0.045 · 60,
Solve for (what):
÷
0.03 · (what) =
0.03
(what)
=
2.7
0.03
90
3% of 90 is equal to 4.5% of 60.
5. (a) (Ideal gas law) Solve for T in pV = nRT .
pV = nRT
pV
nRT
=
nR
nR
pV
=T
nR
(b) (Area of a trapezoid) Solve for b2 in A =
1
h(b1 + b2 ).
2
1
h(b1 + b2 )
2
2
2 1
· A = · h(b1 + b2 )
h
h 2
2A
= b1 + b2
h
A=
2A
− b1 = b1 + b2 − b1
h
2A
− b1 = b2
h
6
0.03 · (what) = 2.7.
(c) Solve for c in abc + 2bc − 3abP = 0.
abc + 2bc − 3abP = 0
abc + 2bc − 3abP + 3abP = 0 + 3abP
abc + 2bc = 3abP
(ab + 2b)c = 3abP
(ab + 2b)c
3abP
=
ab + 2b
ab + 2b
3abP
c=
ab + 2b
(d) Solve for k in ak = b(k + 1).
ak = b(k + 1)
ak = bk + b
ak − bk = b
k(a − b) = b
b
k(a − b)
=
a−b
a−b
b
k=
a−b
6. (a) Solve 3x + 5 > 4(2x − 1). Write your answer using inequality notation, and draw a picture of the
solution set.
3x + 5 >
4(2x − 1)
3x + 5 > 8x
−
4
− 3x
3x
5 > 5x
−
4
+
4
4
9 > 5x
÷
5
5
9
> x
5
9/5
(b) Solve −3x − 4 ≤ −8(x + 5). Write your answer using inequality notation, and draw a picture of the
solution set.
−3x −
4
≤
−8(x + 5)
−3x −
4
≤ −8x
−
40
+ 3x
3x
−4 ≤ −5x
−
40
+
40
40
36
≤ −5x
÷
−5
−5
36
−
≥
x
5
Notice that the inequality “flipped over” when I divided by the negative number −5.
-36/5
7
(c) Solve 0.5(6x − 4) < 3x − 10. Write your answer using inequality notation, and draw a picture of the
solution set.
0.5(6x − 4)
< 3x − 10
3x
−
2 < 3x − 10
− 3x
3x
−2 <
−10
The last statement “−2 < −10” is false. Therefore, the original inequality has no solutions.
(d) Solve −4 < 2(x + 1) < 10. Write your answer using inequality
solution set.
−4 <
2(x + 1)
<
−4 < 2x
+
2 <
− 2
2
−6 < 2x
<
/
2
2
−3 < x
<
-3
notation, and draw a picture of the
10
10
2
8
2
4
4
7. (a) Solve for x: |x − 10| = 5.
|x − 10| = 5
ւ
ց
+
x − 10 = 5
10
10
x
= 15
x − 10
10
x
+
= −5
10
= 5
(b) Solve for x: 5 + |x + 7| = 13.
5
5
−
+
x +
−
x
|x + 7| = 13
5
|x + 7| = 8
ւ
ց
7 = 8
7
7
= 1
−
x + 7 = −8
7
7
x
= −15
(c) Solve for x: |x − 4| = −6.
The absolute value of x − 4 must be greater than or equal to 0; it can’t be negative, so it can’t be equal
to −6. Therefore, there are no solutions.
(d) Solve for x: |2x + 3| = 17.
|2x + 3|
ւ
2x + 3
3
2x
÷ 2
x
−
= 17
ց
= 17
3
= 14
2
= 7
2x + 3 = −17
3
3
2x
= −20
÷ 2
2
x
= −10
−
8
8. (a) Solve for x: |x − 1| < 4.
|x − 1|
the distance from x to 1
<
4
is less than 4
If I move 4 units to the left of 1, I’m at 1 − 4 = −3; if I move 4 units to the right of 1, I’m at 1 + 4 = 5.
4
4
-3
1
5
The solution is −3 < x < 5.
(b) Solve for x: |x + 6| ≥ 3.
Write the inequality as |x − (−6) ≥ 3. (I have to do this because it is |a − b| — not |a + b| — that means
“the distance from a to b”.)
|x − (−6)|
≥
3
the distance from x to −6 is greater than or equal to 3
If I move 3 units to the left of −6, I’m at −6 − 3 = −9; if I move 3 units to the right of −6, I’m at
−6 + 3 = −3.
3
-9
3
-6
-3
The solution is x ≤ −9 or −3 ≤ x.
(c) Solve for x: |x − 17| < −2.
The absolute value of x − 17 is nonegative, so it can’t be less than −2. Therefore, there are no solutions.
(d) Solve for x: |2x − 5| > 13.
First, I’ll divide both sides by 2 to change the 2x to x:
|2x − 5| >
5 2 x − >
2
÷
2
x − 5 >
2
13
13
2
13
2
Now translate the inequality and graph it:
x −
5 2
the distance from x to
>
5
2
9
is greater than
13
2
13
2
5
5
13
13
5
13
units to the left of , I’m at −
= −4; if I move
units to the right of , I’m at
If I move
2
2
2
2
2
2
5 13
+
= 9.
2
2
13/2
-4
13/2
5/2
9
The solution is x < −4 or 9 < x.
(e) Solve for x: | − x − 6| ≤ 1.
I’ll factor out a −1 to change the −x to x. Note that it factors out of the absolute value as | − 1| = 1.
| − x − 6| ≤ 1,
|(−1)(x + 6)| ≤ 1,
| − 1||x + 6| ≤ 1,
1 · |x + 6| ≤ 1,
|x + 6| ≤ 1,
|x − (−6)| ≤ 1.
This inequality can be translated as follows:
|x − (−6)|
≤
1
the distance from x to −6 is less than or equal to 1
If I move 1 unit to the left of −6, I’m at −6 − 1 = −7; if I move 1 unit to the right of −6, I’m at
−6 + 1 = −5.
1
-7
1
-6
-5
The solution is −7 ≤ x ≤ −5.
9. Calvin Butterball has $4.00 in dimes and nickels. The number of dimes is 5 less than twice the number of
nickels. Find the number of dimes and the number of nickels.
Let d be the number of dimes and let n be the number of nickels.
The number of dimes is 5 less than twice the number of nickels, so d = 2n − 5.
There are d dimes and each dime is worth 10 cents, so the dimes are worth 10d cents. There are n
nickels and each nickel is worth 5 cents, so the nickels are worth 5n cents.
Calvin has $4.00 (or 400 cents) all together, so 400 = 10d + 5n.
number of coins
value per coin
total value
dimes
d = 2n − 5
10
10d
nickels
n
5
5n
all together
-
-
400
Plugging d = 2n − 5 into 400 = 10d + 5n gives
400 = 10(2n − 5) + 5n,
Solve for n:
400 = 20n − 50 + 5n,
400 = 25n
50
450 = 25n
÷ 25
25
18 = n
+
10
− 50
50
400 = 25n − 50.
Plugging this into d = 2n − 5 gives d = 36 − 5 = 31.
Calvin has 31 dimes and 18 nickels.
10. Silas Hogwinder has some 13-cent stamps, some 17-cent stamps, and some 40-cent stamps. The number
of 13-cent stamps is 2 less than the number of 17-cent stamps. The number of 40-cent stamps is 1 more than
twice the number of 17-cent stamps. The total value of the stamps is $13.34. Find the number of each type
of stamp that Silas has.
Let n be the number of 17-cent stamps.
The number of 13-cent stamps is 2 less than the number of 17-cent stamps, so the number of 13-cent
stamps is n − 2.
The number of 40-cent stamps is 1 more than twice the number of 17-cent stamps, so the number of
40-cent stamps is 2n + 1.
I put the information into a table:
number of stamps
·
cents per stamp
=
total value (cents)
13-cent stamps
n−2
·
13
=
13(n − 2)
17-cent stamps
n
·
17
=
17n
40-cent stamps
2n + 1
·
40
=
40(2n + 1)
total
1334
The last column says
13(n − 2) + 17n + 40(2n + 1) = 1334.
Solve for n:
13(n − 2) + 17n + 40(2n + 1) = 1334
13n − 26 + 17n + 80n + 40 = 1334
110n + 14 = 1334
110n = 1320
1320
n=
= 12
110
The number of 17-cent stamps is 12, the number of 13-cent stamps is n − 2 = 12 − 2 = 10, and the
number of 40-cent stamps is 2n + 1 = 2(12) + 1 = 25.
11. How many gallons of a 40% alcohol solution and a 30% alcohol solution must be mixed together to make
50 gallons of a 32% alcohol solution?
Let x be the number of gallons of the 40% solution and let y be the number of gallons of the 30%
solution.
There are 50 gallons all together, so x + y = 50.
x gallons of 40% solution contain 40% of x, or 0.4x gallons of alcohol.
y gallons of 30% solution contain 30% of y, or 0.3y gallons of alcohol.
The mixture is 50 gallons of a 32% solution; it contains 32% of 50, or 0.32 · 50 = 16 gallons of alcohol.
Therefore, 0.4x + 0.3y = 16.
gallons of solution
percent alcohol
gallons of alcohol
40% solution
x
40% (= 0.4)
0.4x
30% solution
y
30% (= 0.3)
0.3y
mixutre
50
32% (= 0.32)
16
11
Solve x + y = 50 to get y = 50 − x. Plug this into 0.4x + 0.3y = 16:
0.4x + 0.3(50 − x) = 16,
Solve for x:
0.4x + 15 − 0.3x = 16,
0.1x + 15 = 16.
0.1x + 15 = 16
15
15
0.1x
= 1
÷ 0.1
0.1
x
= 10
−
Plug this into y = 50 − x to get y = 50 − 10 = 40.
You need 10 gallons of the 40% solution and 40 gallons of the 30% solution.
12. How many pounds of dried fruit worth $7 per pound must be mixed with 4 pounds of peppermint ketchup
worth $3.50 per pound to make a mixture worth $5 per pound?
Let x be the number of pounds of dried fruit needed. Fill in the table:
pounds
price per pound
total price
dried fruit
x
7
7x
peppermint ketchup
4
3.5
14
mixture
x+4
5
7x + 14
The last line says:
5(x + 4) = 7x + 14.
Solve for x:
5(x + 4) = 7x + 14,
5x + 20 = 7x + 14,
6 = 2x,
x = 3.
Therefore, 3 pounds of dried fruit are required.
13. Calvin drives down a long straight road in his 1978 Chevette at a constant speed of 30 miles an hour.
Phoebe starts out 50 miles behind him, and follows him at 70 miles an hour in her Porsche. How long will
it take before Phoebe runs Calvin off the road?
Let t be the time (in hours) that Calvin travels before Phoebe catches up with him.
time
·
speed
=
distance
Calvin
t
·
30
=
30t
Phoebe
t
·
70
=
70t
Since Phoebe started 50 miles behind Calvin, she must travel 50 miles more than Calvin:
30t + 50 = 70t.
Solve for t:
30t + 50 = 70t
50 = 40t
5
=t
4
12
Now
5
= 1.25, so it takes Phoebe 1.25 hours (1 hour and 15 minutes) to catch up.
4
14. Leopold drives 6 miles per hour faster than Molly. If they start at the same point and drive in opposite
directions for 4 hours, they will be 272 miles apart. What is Molly’s speed?
Let x be Molly’s speed. Since Leopold drives 6 miles per hour faster than Molly, his speed is x + 6.
time
·
speed
=
distance
Leopold
4
·
x+6
=
4(x + 6)
Molly
4
·
x
=
4x
272
The last column says
4(x + 6) + 4x = 272.
Solve for x:
4(x + 6) + 4x = 272
4x + 24 + 4x = 272
8x + 24 = 272
8x = 248
x = 31
Molly’s speed is 31 miles per hour.
15. Phoebe invests $2000 in two accounts. One pays 6% simple interest, while the other pays 8% simple
interest. At the end of a year, she has earned $128 in interest. How much was invested in each account?
Let x be the amount invested at 6% and let y be the amount invested at 8%.
amount invested
interest rate
interest earned
6% account
x
0.06
0.06x
8% account
y
0.08
0.08y
total
2000
128
I have
x + y = 2000
and 0.06x + 0.08y = 128.
x + y = 2000 gives y = 2000 − x. Plug this into 0.06x + 0.08y = 128 and solve for x:
0.06x + 0.08(2000 − x) = 128
100 · 0.06x + 100 · 0.08(2000 − x) = 100 · 128
6x + 8(2000 − x) = 12800
6x + 16000 − 8x = 12800
−2x + 16000 = 12800
−2x = −3200
x = 1600
Then y = 2000 − 1600 = 400. Thus, $1600 was invested at 6% and $400 was invested at 8%.
13
16. The largest angle in a triangle is 16◦ less than three times the smallest angle. The third angle is 16◦
more than the smallest angle. Find the measures of the angles.
Let S be the smallest angle, let T be the third angle, and let L be the largest angle.
The angles add up to 180◦ , so S + T + L = 180.
The largest angle in a triangle is 16◦ less than three times the smallest angle, so L = 3S − 16.
The third angle is 16◦ more than the smallest angle, so T = 16 + S.
Plug L = 3S − 16 and T = 16 + S into S + T + L = 180 and solve for S:
S + (16 + S) + (3S − 16) = 180
5S = 180
S = 36
Then T = 16 + 36 = 52 and L = 3 · 26 − 16 = 92. The angles are 36◦ , 52◦ , and 92◦ .
17. Multiply the polynomials:
(a) (2x − 3)(x + 4).
2x2
2x2
2x
x
8x
− 3x
+ 5x
− 3
+ 4
− 12
− 12
(b) (3x2 − x − 1)(x − 2).
3x2
3
3x
3x3
−6x2
−
x2
− 7x2
−
x
x
+ 2x
− x
+ x
− 1
− 2
+ 2
+ 2
(c) (2x − 5)(2x + 5).
Using the rule (a − b)(a + b) = a2 − b2 ,
(2x − 5)(2x + 5) = 4x2 − 25.
(d) (3w + 7)2 .
Using the rule (a + b)2 = a2 − 2ab + b2 ,
(3w + 7)2 = 9w2 + 42w + 49.
(e) (1 − 6t)2 .
Using the rule (a − b)2 = a2 − 2ab + b2 ,
(1 − 6t)2 = 1 − 12t + 36t2 .
(f) (2x + y)(x − 3y).
Using FOIL, I get
(2x + y)(x − 3y) = 2x2 − 6xy + xy − 3y 2 = 2x2 − 5xy − 3y 2 .
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(g) (x2 − y 4 )2 .
Using the form (a − b)2 = a2 − 2ab + b2 , I get
(x2 − y 4 )2 = x4 − 2x2 y 4 + y 8 .
(h) (x + y + 1)(x + 2y).
(x+y +1)(x+2y) = (x+y +1)(x)+(x+y +1)(2y) = x2 +xy +x+2xy +2y 2 +2y = x2 +x+3xy +2y 2 +2y.
18. Factor completely:
(a) x2 − 64 = (x − 8)(x + 8).
(b) x2 + 8x + 16 = (x + 4)2 .
(c) 4x2 − 25 = (2x − 5)(2x + 5).
(d) x2 − 5x + 6 = (x − 2)(x − 3).
(e) x2 − 9x − 10 = (x − 10)(x + 1).
(f) 4x2 + 8x − 5 = (2x − 1)(2x + 5).
(g) a4 − 16b4 = (a2 − 4b2 )(a2 + 4b2 ) = (a − 2b)(a + 2b)(a2 + 4b2 ).
(h) x3 − 8y 3 = (x − 2y)(x2 + 2xy + 4y 2 ).
(i) x3 + 5x2 − 16x − 80.
I’ll use factoring by grouping:
x3 +5x2 −16x−80 = (x3 +5x2 )−(16x+80) = x2 (x+5)−16(x+5) = (x2 −16)(x+5) = (x−4)(x+4)(x+5).
(j) x2 y − 3xy + xy 2 − 3y 2 .
I’ll use factoring by grouping:
x2 y−3xy+xy 2 −3y 2 = (x2 y−3xy)+(xy 2 −3y 2 ) = xy(x−3)+y 2 (x−3) = (xy+y 2 )(x−3) = (x+y)(y)(x−3).
(k)
81
− 1.
a2 b2
81
−1=
a2 b2
(l)
9
ab
2
−1 =
2x
y
2
2
9
9
−1
+1 .
ab
ab
4x2
4x
+
+ 1.
y2
y
4x
4x2
+
+1=
y2
y
+2·
2x
+ 12 =
y
2
2x
+1 .
y
(m) a2 − 3a − ab + 3b.
a2 − 3a − ab + 3b = (a2 − 3a) − (ab − 3b) = a(a − 3) − b(a − 3) = (a − b)(a − 3).
(n) x3 + x2 − xy 2 − y 2 .
x3 +x2 −xy 2 −y 2 = (x3 +x2 )−(xy 2 +y 2 ) = x2 (x+1)−y 2 (x+1) = (x+1)(x2 −y 2 ) = (x+1)(x−y)(x+y).
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(o) x2 + xy − 5x − 5y.
x2 + xy − 5x − 5y = x(x + y) − 5(x + y) = (x − 5)(x + y).
(p) 2x3 + 7x2 + 3x.
2x3 + 7x2 + 3x = x(2x2 + 7x + 3) = x(2x + 1)(x + 3).
19. Factor completely.
(a) x3 + 27y 3 .
x3 + 27y 3 = (x + 3y)(x2 − 3xy + 9y 2 ).
(b)
1 3
a − 125b3 .
8
1 3
a − 125b3 =
8
1
1 2 5
2
a − 5b
a + ab + 25b .
2
4
2
(c) 64 − y 3 .
64 − y 3 = (4 − y)(16 + 4y + y 2 ).
(d) ab + 6b + a2 + 6a.
ab + 6b + a2 + 6a = (ab + 6b) + (a2 + 6a) = b(a + 6) + a(a + 6) = (b + a)(a + 6).
(e) x3 − 3x2 − 4x + 12.
x3 − 3x2 − 4x + 12 = (x3 − 3x2 ) − (4x − 12) = x2 (x − 3) − 4(x − 3) = (x − 3)(x2 − 4) = (x − 3)(x − 2)(x + 2).
(f) x2 y − 2xy 2 + x − 2y.
x2 y − 2xy 2 + x − 2y = (x2 y − 2xy 2 ) + (x − 2y) = xy(x − 2y) + (x − 2y) = (x − 2y)(xy + 1).
20. (a) Divide x4 + x3 + 1 by x2 − 1 using long division.
x4 + x3 + 1
x+2
= (x2 + x + 1) + 2
.
x2 − 1
x −1
(b) Compute the quotient and remainder when 2x4 + 3x2 + 4x is divided by 2x2 .
2x4 + 3x2 + 4x
2x4
3x2
4x
3 2
=
+
+ 2 = x2 + + .
2
2
2
2x
2x
2x
2x
2 x
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Since there’s only one term on the bottom, it’s easier to break the fraction up into pieces than to do
the long division.
(c) Compute the quotient and remainder when 2x2 + 5xy − 3y 2 is divided by x + 3y.
2x2 + 5xy − 3y 2
= 2x − y.
x + 3y
(d) Factor x3 − 2x2 − 11x + 12 completely, given that x − 4 is one of the factors.
Divide x3 − 2x2 − 11x + 12 by x − 4:
x 2 + 2x - 3
x-4
x 3 - 2 x 2 - 11x + 12
x3 - 4x2
2 x 2 - 11x
2 x 2 - 8x
-3x + 12
-3x + 12
Thus,
x3 − 2x2 − 11x + 12 = (x − 4)(x2 + 2x − 3) = (x − 4)(x + 3)(x − 1).
If all the water in the Mississippi River were placed in a tank on the moon, there would be a big ditch running
down the middle of the United States. - National Lampoon
c 2008 by Bruce Ikenaga
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