11
5B. SPECIAL PRODUCTS
5b
Special Products
Special Forms
In this section we revisit two special product forms that we learned in Chapter
5, the first of which was squaring a binomial.
Squaring a binomial. Here are two earlier rules for squaring a binomial.
1. (a + b)2 = a2 + 2ab + b2
2. (a − b)2 = a2 − 2ab + b2
Now, because factoring is “unmultiplying,” it should be a simple matter to
reverse the process of Example ??.
You Try It!
EXAMPLE 1. Factor each of the following trinomials:
a) x2 + 10x + 25
b) 4y 2 − 12y + 9
c) 16a2 − 24ab + 9b2
Solution: Because of the work already done in Example ??, it is a simple task
to factor each of these trinomials.
a) x2 + 10x + 25 = (x + 5)2
b) 4y 2 − 12y + 9 = (2y − 3)2
c) 16a2 − 24ab + 9b2 = (4a − 3b)2
!
You Try It!
EXAMPLE 2. Factor each of the following trinomials:
a) 9x2 − 42x + 49
b) 49a2 + 70ab + 25b2
c) 4x2 − 37x + 9
Solution: Note that the first and last terms of each trinomial are perfect
squares.
a) In the trinomial 9x2 − 42x + 49, note that (3x)2 = 9x2 and 72 = 49. Hence,
the first and last terms are perfect squares. Taking the square roots, we
suspect that 9x2 − 42x + 49 factors as follows:
9x2 − 42x + 49 = (3x − 7)2
However, we must check to see if the middle term is correct. Multiply 3x
and 7, then double: 2(3x)(7) = 42x. Thus, the middle term is correct and
we have the correct factorization of 9x2 − 42x + 49.
12
MODULE 5. FACTORING
b) In the trinomial 49a2 +70ab+25b2, note that (7a)2 = 49a2 and (5b)2 = 25b2.
Hence, the first and last terms are perfect squares. Taking the square roots,
we suspect that 49a2 + 70ab + 25b2 factors as follows:
49a2 + 70ab + 25b2 = (7a + 5b)2
However, we must check to see if the middle term is correct. Multiply 7a
and 5b, then double: 2(7a)(5b) = 70ab. Thus, the middle term is correct
and we have the correct factorization of 49a2 + 70ab + 25b2 .
c) In the trinomial 4x2 − 37x + 9, note that (2x)2 = 4x2 and (3)2 = 9. Hence,
the first and last terms are perfect squares. Taking the square roots, we
suspect that 4x2 − 37x + 9 factors as follows:
4x2 − 37x + 9 = (2x − 3)2
However, we must check to see if the middle term is correct. Multiply 2x
and 3, then double: 2(2x)(3) = 12x. However, this is not the middle term
of 4x2 − 37x + 9, so this factorization is incorrect! We must find another
way to factor this trinomial.
Comparing 4x2 − 37x + 9 with ax2 + bx + c, we need a pair of integers
whose product is ac = 36 and whose sum is b = −37. The integer pair −1
and −36 comes to mind. Replace the middle term as a sum of like terms
using this ordered pair.
4x2 − 37x + 9 = 4x2 − x − 36x + 9
= x(4x − 1) − 9(4x − 1)
= (x − 9)(4x − 1)
−37x = −x − 36x.
Factor by grouping.
Factor out 4x − 1.
This example clearly demonstrates how important it is to check your middle
term.
!
The first rule of factoring. The first step to perform in any factoring
problem is factor out the GCF.
You Try It!
EXAMPLE 3. Factor each of the following trinomials:
a) 2x3 y + 12x2 y 2 + 18xy 3
Solution: First factor out the GCF.
b) −4x5 + 32x4 − 64x3
13
5B. SPECIAL PRODUCTS
a) In the trinomial 2x3 y + 12x2 y 2 + 18xy 3 , we note that the GCF of 2x3 y,
12x2 y 2 , and 18xy 3 is 2xy. We first factor out 2xy.
2x3 y + 12x2 y 2 + 18xy 3 = 2xy(x2 + 6xy + 9y 2 )
We now note that the first and last terms of the resulting trinomial factor
are perfect squares, so we take their square roots and try the following:
= 2xy(x + 3y)2
Of course, we must check that our middle term is correct. Because 2(x)(3y) =
6xy matches the middle term of x2 + 6xy + 9y 2, we do have a perfect square
trinomial and our result is correct.
b) In the trinomial −4x5 + 32x4 − 64x3 , we note that the GCF of 4x5 , 32x4 ,
and 64x3 is 4x3 . We first factor out 4x3 .
−4x5 + 32x4 − 64x3 = 4x3 (−x2 + 8x − 16)
However, the first and third terms of −x2 + 8x − 16 are negative, and thus
are not perfect squares. Let’s begin again, this time factoring out −4x3 .
−4x5 + 32x4 − 64x3 = −4x3 (x2 − 8x + 16)
This time the first and third terms of x2 − 8x + 16 are perfect squares. We
take their square roots and write:
= −4x3 (x − 4)2
We must check that our middle term is correct. Because 2(x)(4) = 8x, we
do have a perfect square trinomial and our result is correct.
!
The Difference of Squares
The second special product form we learned in Chapter 5 was the difference of
squares.
The difference of squares. Here is the difference of squares rule.
(a + b)(a − b) = a2 − b2
Because factoring is “unmultiplying,” is can be a simple matter to reverse
the process.
You Try It!
EXAMPLE 4. Factor each of the following:
14
MODULE 5. FACTORING
a) 9x2 − 25
b) a6 − 4b6
Solution: Because of the work already done in previous examples, it is now a
matter to factor (or “unmultiply”) each of these problems.
a) 9x2 − 25 = (3x + 5)(3x − 5)
b) a6 − 4b6 = (a3 − 2b3 )(a3 + 2b3 )
In each case, note how we took the square roots of each term, then separated
one set with a plus sign and the other with a minus sign. Because of the
commutative property of multiplication, it does not matter which one you
make plus and which one you make minus.
!
Remember the first rule of factoring.
The first rule of factoring. The first step to perform in any factoring
problem is factor out the GCF.
You Try It!
EXAMPLE 5. Factor: x3 − 9x
Solution: In x3 − 9x, the GCF of x3 and 9x is x. Factor out x.
x3 − 9x = x(x2 − 9)
Note that x2 − 9 is now the difference of two perfect squares. Take the square
roots of x2 and 9, which are x and 3, then separate one set with a plus sign
and the other set with a minus sign.
= x(x + 3)(x − 3)
!
Factoring Completely
Sometimes after one pass at factoring, factors remain that can be factored
further. You must continue to factor in this case.
You Try It!
EXAMPLE 6. Factor: x4 − 16
15
5B. SPECIAL PRODUCTS
Solution: In x4 − 16, we have the difference of two squares: (x2 )2 = x4 and
42 = 16. First, we take the square roots, then separate one set with a plus sign
and the other set with a minus sign.
x4 − 16 = (x2 + 4)(x2 − 4)
Note that x2 +4 is the sum of two squares and does not factor further. However,
x2 − 4 is the difference of two squares. Take the square roots, x and 2, then
separate one set with a plus sign and the other set with a minus sign.
= (x2 + 4)(x + 2)(x − 2)
We cannot factor further.
!
Revisiting Factoring by Grouping
We usually factor a four-term expression by grouping.
You Try It!
EXAMPLE 7. Factor: 2x3 + x2 − 50x − 25
Solution: We factor by grouping. Factor an x2 out of the first two terms and
a −25 out of the second two terms.
2x3 + x2 − 50x − 25 = x2 (2x + 1) − 25(2x + 1)
Now we can factor out a 2x + 1.
= (x2 − 25)(2x + 1)
We’re still not done because x2 − 25 is the difference of two squares and can
be factored as follows:
= (x + 5)(x − 5)(2x + 1)
!
Factoring Trinomials I
In this section we concentrate on learning how to factor trinomials having the
form ax2 + bx + c which are not perfect squares. We must first be able to
identify the coefficients a, b, and c.
16
MODULE 5. FACTORING
You Try It!
EXAMPLE 8. Compare 3 − 7x + 9x2 with ax2 + bx + c and identify the
coefficients a, b, and c.
Solution: Note that ax2 + bx + c is arranged in descending powers of x, but
3 − 7x + 9x2 is arranged in descending powers of x. The first task is to arrange
3 − 7x + 9x2 in descending powers of x, then align it with ax2 + bx + c for
comparison.
ax2 + bx +c
9x2 − 7x+3
Comparing, a = 9, b = −7, and c = 3.
!
You Try It!
EXAMPLE 9. Compare −5 + x2 + 3x with ax2 + bx + c and identify the
coefficients a, b, and c.
Solution: First, arrange −5 + x2 + 3x in descending powers of x, then align
it with ax2 + bx + c for comparison. Note that the understood coefficient of x2
is 1.
ax2 + bx +c
1x2 + 3x−5
Comparing, a = 1, b = 3, and c = −5.
!
The ac-Method
We are now going to revie a technique called the ac-method (or ac-test ) for
factoring trinomials of the form ax2 + bx + c. We begin with a simple example.
You Try It!
EXAMPLE 10. Factor: x2 + 8x − 48.
Solution: We proceed as follows:
1. Compare x2 + 8x − 48 with ax2 + bx + c and identify a = 1, b = 8, and
c = −48. Note that the leading coefficient is a = 1.
2. Calculate ac. Note that ac = (1)(−48), so ac = −48.
17
5B. SPECIAL PRODUCTS
3. List all integer pairs whose product is ac = −48.
1, −48
2, −24
3, −16
4, −12
6, −8
−1, 48
−2, 24
−3, 16
−4, 12
−6, 8
4. Circle the ordered pair whose sum is b = 8.
1, −48
2, −24
3, −16
4, −12
6, −8
−1, 48
−2, 24
−3, 16
−4, 12
−6, 8
5. “Drop” this ordered pair in place.
x2 + 8x − 48 = (x − 4)(x + 12)
6. Use the FOIL shortcut to mentally check your answer. To multiply (x −
4)(x + 12), use these steps:
• Multiply the terms in the “First” positions: x2 .
• Multiply the terms in the “Outer” and “Inner” positions and combine the results mentally: 12x − 4x = 8x.
• Multiply the terms in the “Last” positions: −48.
That is:
(x − 4)(x + 12) =
F
x2
O
I
+ 12x − 4x −
L
48
Combining like terms, (x− 4)(x+ 12) = x2 + 8x− 48, which is the original
trinomial, so our solution checks.
!
Speeding Things Up a Bit
Some readers might already be asking “Do I really have to list all of those
ordered pairs if I already see the pair I need?” The answer is “No!” If you see
the pair you need, drop it in place and you are done.
You Try It!
EXAMPLE 11. Factor: x2 − 5x − 24.
18
MODULE 5. FACTORING
Solution: Compare x2 −5x−24 with ax2 +bx+c and note that a = 1, b = −5,
and c = −24. Calculate ac = (1)(−24), so ac = −24. Now can you think of an
integer pair whose product is ac = −24 and whose sum is b = −5? For some,
the pair just pops into their head: −8 and 3. “Drop” the pair in place and you
are done.
x2 − 5x − 24 = (x − 8)(x + 3)
Use the FOIL shortcut to check your answer.
(x − 8)(x + 3) =
F
x2
+
O
I
3x − 8x −
L
24
Combining like terms, (x − 8)(x + 3) = x2 − 5x − 24, the original trinomial. Our
solution checks. Note that if you combine the “Outer” and “Inner” products
mentally, the check goes even faster.
!
You Try It!
EXAMPLE 12. Factor: x2 − 9x − 36.
Solution: Compare x2 −9x−36 with ax2 +bx+c and note that a = 1, b = −9,
and c = −36. Calculate ac = (1)(−36), so ac = −36. Start listing the integer
pairs whose product is ac = −36, but be mindful that you need an integer pair
whose sum is b = −9.
1, −36
2, −18
3, −12
Note how we ceased listing ordered pairs the moment we found the pair we
needed. “Drop” the circled pair in place.
x2 − 9x − 36 = (x + 3)(x − 12)
Use the FOIL shortcut to check your answer.
(x + 3)(x − 12) =
F
x2
−
O
12x +
I
3x −
L
36
Combining like terms, (x+3)(x−12) = x2 −9x−36, the original trinomial. Our
solution checks. Note that if you combine the “Outer” and “Inner” products
mentally, the check goes even faster.
!
19
5B. SPECIAL PRODUCTS
Factoring Trinomials II
In this section we continue to factor trinomials of the form ax2 + bx + c. In
the last section, all of our examples had a = 1, and we were able to “Drop in
place” our circled integer pair. However, in this section, a != 1, and we’ll soon
see that we will not be able to use the “Drop in place” technique. However,
readers will be pleased to learn that the ac-method will still apply, only with
a slightly different twist.
You Try It!
EXAMPLE 13. Factor: 2x2 − 7x − 15.
Solution: We proceed as follows:
1. Compare 2x2 − 7x − 15 with ax2 + bx + c and identify a = 2, b = −7,
and c = −15. Note that the leading coefficient is a = 2, so this case is
different from all of the cases discussed in Section 5b.
2. Calculate ac. Note that ac = (2)(−15), so ac = −30.
3. List all integer pairs whose product is ac = −30.
1, −30
2, −15
3, −10
5, −6
−1, 30
−2, 15
−3, 10
−5, 6
4. Circle the ordered pair whose sum is b = −7.
1, −30
2, −15
3, −10
5, −6
−1, 30
−2, 15
−3, 10
−5, 6
5. Note that if we “drop in place” our circled ordered pair, (x + 3)(x − 10) !=
2x2 − 7x − 15. Right off the bat, the product of the terms in the “First”
position does not equal 2x2 . Instead, we break up the middle term of
2x2 − 7x − 15 into a sum of like terms using our circled pair of integers
3 and −10.
2x2 −7x − 15 = 2x2 +3x − 10x − 15
Now we factor by grouping. Factor an x out of the first two terms and a
−5 out of the second two terms.
= x (2x + 3) − 5 (2x + 3)
Now we can factor out (2x + 3).
= (x − 5)(2x + 3)
20
MODULE 5. FACTORING
6. Use the FOIL shortcut to mentally check your answer. To multiply
(x − 5)(2x + 3), use these steps:
• Multiply the terms in the “First” positions: 2x2 .
• Multiply the terms in the “Outer” and “Inner” positions and combine the results mentally: 3x − 10x = −7x.
• Multiply the terms in the “Last” positions: −15.
That is:
(x − 5)(2x + 3) =
F
2x2
O
+ 3x −
I
L
10x − 15
Combining like terms, (x − 5)(2x + 3) = 2x2 − 7x − 15, which is the
original trinomial, so our solution checks.
!
Speeding Things Up a Bit
Some readers might already be asking “Do I really have to list all of those
ordered pairs if I already see the pair I need?” The answer is “No!” If you see
the pair you need, use it to break up the middle term of the trinomial as a sum
of like terms.
You Try It!
EXAMPLE 14. Factor: 3x2 − 7x − 6.
Solution: Compare 3x2 −7x−6 with ax2 +bx+c and note that a = 3, b = −7,
and c = −6. Calculate ac = (3)(−6), so ac = −18. Now can you think of an
integer pair whose product is ac = −18 and whose sum is b = −7? For some,
the pair just pops into their head: 2 and −9. Break up the middle term into a
sum of like terms using the pair 2 and −9.
3x2 −7x − 6
= 3x2 +2x − 9x − 6
−7x = 2x − 9x.
= x (3x + 2) − 3 (3x + 2)
= (x − 3)(3x + 2)
Factor by grouping.
Factor out (3x + 2).
Use the FOIL shortcut to check your answer.
(x − 3)(3x + 2) =
F
3x2
+
O
I
2x − 9x −
L
6
Combining like terms, (x−3)(3x+2) = 3x2 −7x−6, the original trinomial. Our
solution checks. Note that if you combine the “Outer” and “Inner” products
mentally, the check goes even faster.
21
5B. SPECIAL PRODUCTS
!
First rule of factoring. The first step in factoring any polynomial is to factor
out the greatest common factor.
You Try It!
EXAMPLE 15. Factor: 30x3 − 21x2 − 18x.
Solution: Note that the GCF of 30x3 , 21x2 , and 18x is 3x. Factor out this
GCF.
30x3 − 21x2 − 18x = 3x · 10x2 − 3x · 7x − 3x · 6
= 3x(10x2 − 7x − 6)
Next, compare 10x2 − 7x − 6 with ax2 + bx + c and note that a = 10, b = −7,
and c = −6. Start listing the integer pairs whose product is ac = −60, but be
mindful that you need an integer pair whose sum is b = −7.
1, −60
2, −30
3, −20
4, −15
5, −12
Break up the middle term into a sum of like terms using our circled pair.
3x(10x2 − 7x − 6)
= 3x(10x2 + 5x − 12x − 18)
!
"
= 3x 5x(2x + 1) − 6(2x + 1)
= 3x(5x − 6)(2x + 1)
−7x = 5x − 12x.
Factor by grouping.
Factor out a 2x + 1.
Hence, 30x3 − 21x2 − 18x = 3x(5x − 6)(2x + 1).
Check: First, use the FOIL shortcut to multiply the two binomial factors,
then distribute the monomial factor.
3x(5x − 6)(2x + 1) = 3x(10x2 − 7x − 6)
3
2
= 30x − 21x − 18x
Apply the FOIL shortcut.
Distribute the 3x.
Because this is the original polynomial, the solution checks.
!