440
6.5
CHAPTER 6. FACTORING
Factoring 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
Perfect Square Trinomials
To square a binomial such as (a + b)2 , proceed as follows:
1. Square the first term: a2
2. Multiply the first and second term, then double: 2ab
3. Square the last term: b2
You Try It!
Expand: (5a + 2b)2
EXAMPLE 1. Expand: (2x + 3y)2
Solution: Using the pattern (a+ b)2 = a2 + 2ab + b2 , we can expand (2x+ 3y)2
as follows:
(2x + 3y)2 = (2x)2 + 2(2x)(3y) + (3y)2
= 4x2 + 6xy + 9y 2
Answer: 25a2 + 20ab + 4b2
Note how we square the first and second terms, then produce the middle term
of our answer by multiplying the first and second terms and doubling.
You Try It!
Expand: (2s3 − 7t)2
EXAMPLE 2. Expand: (3u2 − 5v 2 )2
Solution: Using the pattern (a−b)2 = a2 −2ab+b2, we can expand (3u2 −5v 2 )2
as follows:
(3u2 − 5v 2 )2 = (3u2 )2 − 2(3u2 )(5v 2 ) + (5v 2 )2
= 9u4 − 30u2 v 2 + 25v 4
441
6.5. FACTORING SPECIAL FORMS
Note that the sign of the middle term is negative this time. The first and last
terms are still positive because we are squaring.
Answer: 4s6 − 28s3 t + 49t2
Once you’ve squared a few binomials, it’s time to do all of the work in your
head. (i) Square the first term; (ii) multiply the first and second term and
double the result; and (iii) square the second term.
You Try It!
Expand: (5x4 − 3)2
EXAMPLE 3. Expand each of the following:
a) (2y − 3)2
b) (4a − 3b)2
c) (x3 + 5)2
Solution: Using the pattern (a ± b)2 = a2 ± 2ab + b2 , we expand each
binomially mentally, writing down the answer without any intermediate steps.
a) (2y − 3)2 = 4y 2 − 12y + 9
b) (4a − 3b)2 = 16a2 − 24ab + 9b2
c) (x3 + 5)2 = x6 + 10x3 + 25
Answer: 25x8 − 30x4 + 9
Now, because factoring is “unmultiplying,” it should be a simple matter to
reverse the process of Example 3.
You Try It!
EXAMPLE 4. Factor each of the following trinomials:
a) 4y 2 − 12y + 9
b) 16a2 − 24ab + 9b2
Factor: 25x8 − 30x4 + 9
c) x6 + 10x3 + 25
Solution: Because of the work already done in Example 3, it is a simple task
to factor each of these trinomials.
a) 4y 2 − 12y + 9 = (2y − 3)2
b) 16a2 − 24ab + 9b2 = (4a − 3b)2
c) x6 + 10x3 + 25 = (x3 + 5)2
Answer: (5x4 − 3)2
442
CHAPTER 6. FACTORING
Each of the trinomials in Example 4 is an example of a perfect square trinomial.
Perfect square trinomial. If a trinomial a2 + 2ab + b2 is the square of a
binomial, as in (a + b)2 , then the trinomial is called a perfect square trinomial.
So, how does one recognize a perfect square trinomial? If the first and last
terms of a trinomial are perfect squares, then you should suspect that you may
be dealing with a perfect square trinomial. However, you also have to have the
correct middle term in order to have a perfect square trinomial.
You Try It!
Factor: 16x2 + 72x + 81
EXAMPLE 5. 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.
List of Squares
n
n2
0
0
1
1
2
4
3
9
4
16
5
25
6
36
7
49
8
64
9
81
10
100
11
121
12
144
13
169
14
196
15
225
16
256
17
289
18
324
19
361
20
400
21
441
22
484
23
529
24
576
25
625
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
therefore
9x2 − 42x + 49 = (3x − 7)2 .
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 therefore
49a2 + 70ab + 25b2 = (7a + 5b)2 .
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
443
6.5. FACTORING SPECIAL FORMS
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 the middle
term.
Answer: (4x + 9)2
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 6. Factor each of the following trinomials:
a) 2x3 y + 12x2 y 2 + 18xy 3
b) −4x5 + 32x4 − 64x3
Solution: Remember, first factor out the GCF.
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 factors as follows.
= 2xy(x + 3y)2
Of course, the last factorization is correct only if the 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.
Factor: −4x3 − 24x2 − 36x
444
CHAPTER 6. FACTORING
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
Again, this last factorization is correct only if the middle term is correct.
Because 2(x)(4) = 8x, we do have a perfect square trinomial and our result
is correct.
Answer: −4x(x + 3)2
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
If you are multiplying two binomials which have the exact same terms in
the “First” positions and the exact same terms in the “Last” positions, but
one set is separated by a plus sign while the other set is separated by a minus
sign, then multiply as follows:
1. Square the first term: a2
2. Square the second term: b2
3. Place a minus sign between the two squares.
445
6.5. FACTORING SPECIAL FORMS
You Try It!
EXAMPLE 7. Expand each of the following:
Expand: (4x − 3y)(4x + 3y)
b) (a3 − 2b3 )(a3 + 2b3 )
a) (3x + 5)(3x − 5)
Solution: We apply the difference of squares pattern to expand each of the
given problems.
a) In (3x + 5)(3x − 5), we have the exact same terms in the “First” and
“Last” positions, with the first set separated by a plus sign and the second
set separated by a minus sign.
a) Square the first term: (3x)2 = 9x2
b) Square the second term: 52 = 25
c) Place a minus sign between the two squares.
Hence:
(3x + 5)(3x − 5) = 9x2 − 25
b) In (a3 − 2b3 )(a3 + 2b3 ), we have the exact same terms in the “First” and
“Last” positions, with the first set separated by a minus sign and the second
set separated by a plus sign.
a) Square the first term: (a3 )2 = a6
b) Square the second term: (2b3 )2 = 4b6
c) Place a minus sign between the two squares.
Hence:
(a3 − 2b3 )(a3 + 2b3 ) = a6 − 4b6
Answer: 16x2 − 9y 2
Because factoring is “unmultiplying,” is should be a simple matter to reverse
the process of Example 7.
You Try It!
EXAMPLE 8. Factor each of the following:
a) 9x2 − 25
b) a6 − 4b6
Solution: Because of the work already done in Example 7, it is a simple
matter to factor (or “unmultiply”) each of these problems.
a) 9x2 − 25 = (3x + 5)(3x − 5)
b) a6 − 4b6 = (a3 − 2b3 )(a3 + 2b3 )
Factor: 81x2 − 49
446
Answer: (9x + 7)(9x − 7)
CHAPTER 6. FACTORING
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.
Always 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!
Factor: 4x4 − 16x2
EXAMPLE 9. 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)
Answer: 4x2 (x + 2)(x − 2)
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!
Factor: x4 − 81
EXAMPLE 10. Factor: x4 − 16
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)
447
6.5. FACTORING SPECIAL FORMS
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)
Answer:
(x2 + 9)(x + 3)(x − 3)
Done. We cannot factor further.
Nonlinear Equations Revisited
Remember, if an equation is nonlinear, the first step is to make one side equal
to zero by moving all terms to one side of the equation. Once you’ve completed
this important first step, factor and apply the zero product property to find
the solutions.
You Try It!
EXAMPLE 11. Solve for x:
25x2 = 169
Solve for x:
16x2 = 121
Solution: Make one side equal to zero, factor, then apply the zero product
property.
25x2 = 169
2
25x − 169 = 0
Original equation.
Subtract 169 from both sides.
Note that we have two perfect squares separated by a minus sign. This is the
difference of squares pattern. Take the square roots, making one term plus and
one term minus.
(5x + 13)(5x − 13) = 0
Use difference of squares to factor.
Use the zero product property to complete the solution, setting each factor
equal to zero and solving the resulting equations.
5x + 13 = 0
13
x=−
5
or
5x − 13 = 0
13
x=
5
Hence, the solutions of 25x2 = 169 are x = −13/5 and x = 13/5. We encourage
readers to check each of these solutions.
Answer: −11/4, 11/4
448
CHAPTER 6. FACTORING
You Try It!
Solve for x:
25x2 = 80x − 64
EXAMPLE 12. Solve for x:
Solution: Make one side equal to zero, factor, then apply the zero product
property.
49x2 + 81 = 126x
One can also argue that the
only number whose square
is zero is the number zero.
Hence, one can go directly
from
49x2 + 81 = 126x
2
49x − 126x + 81 = 0
Original equation.
Subtract 126x from both sides.
Note that the first and last terms of the trinomial are perfect squares. Hence,
it make sense to try and factor as a perfect square trinomial, taking the square
roots of the first and last terms.
(7x − 9)2 = 0
(7x − 9)2 = 0
Factor as a perfect square trinomial.
to
7x − 9 = 0.
Hence, the only solution of
49x2 + 81 = 126x is x = 9/7.
Of course, be sure to check the middle term. Because −2(7x)(9) = −126x, the
middle term is correct. Because (7x − 9)2 = (7x − 9)(7x − 9), we can use the
zero product property to set each factor equal to zero and solve the resulting
equations.
7x − 9 = 0
9
x=
7
Answer: 8/5
or
7x − 9 = 0
9
x=
7
Hence, the only solution of 49x2 + 81 = 126x is x = 9/7. We encourage readers
to check this solution.
You Try It!
Solve for x:
5x3 + 36 = x2 + 180x
EXAMPLE 13. Solve for x:
2x3 + 3x2 = 50x + 75
Solution: Make one side equal to zero, factor, then apply the zero product
property.
2x3 + 3x2 = 50x + 75
3
2
2x + 3x − 50x − 75 = 0
Original equation.
Make one side zero.
This is a four-term expression, so we try factoring by grouping. Factor x2 out
of the first two terms, and −25 out of the second two terms.
x2 (2x + 3) − 25(2x + 3) = 0
2
(x − 25)(2x + 3) = 0
Factor by grouping.
Factor out 2x + 3.
449
6.5. FACTORING SPECIAL FORMS
Complete the factorization by using the difference of squares to factor x2 − 25.
(x + 5)(x − 5)(2x + 3) = 0
Use difference of squares to factor.
Finally, use the zero product property. Set each factor equal to zero and solve
for x.
x+5=0
or
x = −5
x−5=0
or
2x + 3 = 0
x=5
x=−
3
2
Hence, the solutions of 2x3 + 3x2 = 50x + 75 are x = −5, x = 5, and x = −3/2.
We encourage readers to check each of these solutions.
Answer: −6, 6, 1/5
Let’s solve another nonlinear equation, matching the algebraic and graphical solutions.
You Try It!
EXAMPLE 14. Solve the equation x3 = 4x, both algebraically and graphically, then compare your answers.
Solution: Note that we have a power of x larger than one, so the equation
x3 = 4x is nonlinear. Make one side zero and factor.
x3 = 4x
Original equation.
3
Nonlinear. Make one side zero.
2
Factor out GCF.
Apply difference of squares.
x − 4x = 0
x(x − 4) = 0
x(x + 2)(x − 2) = 0
Note that we now have a product of three factors that equals zero. The zero
product property says that at least one of these factors must equal zero.
x=0
or
x+2=0
x = −2
or
x−2=0
x=2
Hence, the solutions of x3 = 4x are x = 0, x = −2, and x = 2.
Graphical solution. Load y = x3 and y = 4x into Y1 and Y2 in the
Y= menu of your calculator. Select 6:ZStandard from the ZOOM menu to
produce the graph in Figure 6.26.
Although the image in Figure 6.26 shows all three points of intersection,
adjusting the WINDOW parameters as shown in Figure 6.27, then pressing
the GRAPH button will produce a nicer view of the points of intersection, as
shown in the figure on the right in Figures 6.27.
Solve the equation x3 = 16x
both algebraically and
graphically, then compare
your answers.
450
CHAPTER 6. FACTORING
Figure 6.26: Sketching y = x3 and y = 4x.
Figure 6.27: Adjusting the viewing window.
Use the 5:intersect tool from the CALC menu to find the three points of
intersection. Press the ENTER key in response to “First curve,” then press
ENTER again in response to “Second curve,” then use the left-arrow key to
move your cursor close to the leftmost point of intersection and press ENTER
in response to “Guess.” The result is shown in the first image on the left in
Figure 6.28. Repeat the process to find the remaining points of intersection.
The results are shown in the last two images in Figure 6.28.
Figure 6.28: Finding the points of intersection.
Thus, the graphical solutions are x = −2, x = 0, and x = 2.
Reporting the solution on your homework: Duplicate the image in your
calculator’s viewing window on your homework page. Use a ruler to draw all
lines, but freehand any curves.
• Label the horizontal and vertical axes with x and y, respectively (see
Figure 6.29).
• Place your WINDOW parameters at the end of each axis (see Figure 6.29).
• Label the graph with its equation (see Figure 6.29).
451
6.5. FACTORING SPECIAL FORMS
• Drop dashed vertical lines through each x-intercept. Shade and label the
x-values of the points where the dashed vertical line crosses the x-axis.
These are the solutions of the equation x3 = 4x (see Figure 6.29).
y
15
y = x3
y = 4x
Answer: −4, 0, 4
−5
0
−2
2
5
x
−15
Figure 6.29: Reporting your graphical solution on your homework.
Finally, note that the graphical solutions x = −2, x = 0, and x = 2 match our
algebraic solutions exactly.
452
CHAPTER 6. FACTORING
❧ ❧ ❧
Exercises
❧ ❧ ❧
In Exercises 1-8, expand each of the given expressions.
1. (8r − 3t)2
5. (s3 − 9)2
2. (6a + c)2
6. (w3 + 7)2
3. (4a + 7b)2
7. (s2 + 6t2 )2
4. (4s + t)2
8. (7u2 − 2w2 )2
In Exercises 9-28, factor each of the given expressions.
9. 25s2 + 60st + 36t2
19. 49r6 + 112r3 + 64
10. 9u2 + 24uv + 16v 2
20. a6 − 16a3 + 64
11. 36v 2 − 60vw + 25w2
21. 5s3 t − 20s2 t2 + 20st3
12. 49b2 − 42bc + 9c2
22. 12r3 t − 12r2 t2 + 3rt3
13. a4 + 18a2 b2 + 81b4
23. 8a3 c + 8a2 c2 + 2ac3
14. 64u4 − 144u2 w2 + 81w4
24. 18x3 z − 60x2 z 2 + 50xz 3
15. 49s4 − 28s2 t2 + 4t4
25. −48b3 + 120b2 − 75b
16. 4a4 − 12a2 c2 + 9c4
26. −45c3 + 120c2 − 80c
17. 49b6 − 112b3 + 64
27. −5u5 − 30u4 − 45u3
18. 25x6 − 10x3 + 1
28. −12z 5 − 36z 4 − 27z 3
In Exercises 29-36, expand each of the given expressions.
29. (21c + 16)(21c − 16)
33. (3y 4 + 23z 4)(3y 4 − 23z 4)
30. (19t + 7)(19t − 7)
34. (5x3 + z 3 )(5x3 − z 3 )
31. (5x + 19z)(5x − 19z)
35. (8r5 + 19s5 )(8r5 − 19s5 )
32. (11u + 5w)(11u − 5w)
36. (3u3 + 16v 3 )(3u3 − 16v 3 )
453
6.5. FACTORING SPECIAL FORMS
In Exercises 37-60, factor each of the given expressions.
37. 361x2 − 529
49. 72y 5 − 242y 3
38. 9b2 − 25
50. 75y 5 − 147y 3
39. 16v 2 − 169
51. 1444a3 b − 324ab3
40. 81r2 − 169
52. 12b3 c − 1875bc3
41. 169x2 − 576y 2
53. 576x3 z − 1156xz 3
42. 100y 2 − 81z 2
54. 192u3 v − 507uv 3
43. 529r2 − 289s2
55. 576t4 − 4t2
44. 49a2 − 144b2
56. 4z 5 − 256z 3
45. 49r6 − 256t6
57. 81x4 − 256
46. 361x10 − 484z 10
58. 81x4 − 1
47. 36u10 − 25w10
59. 81x4 − 16
48. a6 − 81c6
60. x4 − 1
In Exercises 61-68, factor each of the given expressions completely.
61. z 3 + z 2 − 9z − 9
65. r3 − 3r2 t − 25rt2 + 75t3
62. 3u3 + u2 − 48u − 16
66. 2b3 − 3b2 c − 50bc2 + 75c3
63. x3 − 2x2 y − xy 2 + 2y 3
67. 2x3 + x2 − 32x − 16
64. x3 + 2x2 z − 4xz 2 − 8z 3
68. r3 − 2r2 − r + 2
In Exercises 69-80, solve each of the given equations for x.
69. 2x3 + 7x2 = 72x + 252
75. 16x2 = 169
70. 2x3 + 7x2 = 32x + 112
76. 289x2 = 4
71. x3 + 5x2 = 64x + 320
77. 9x2 = 25
72. x3 + 4x2 = 49x + 196
78. 144x2 = 121
73. 144x2 + 121 = 264x
79. 256x2 + 361 = −608x
74. 361x2 + 529 = 874x
80. 16x2 + 289 = −136x
454
CHAPTER 6. FACTORING
In Exercises 81-84, perform each of the following tasks:
i) Use a strictly algebraic technique to solve the given equation.
ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the
results found using graphing calculator as shown in Example 12.
81. x3 = x
83. 4x3 = x
82. x3 = 9x
84. 9x3 = x
❧ ❧ ❧
Answers
❧ ❧ ❧
1. 64r2 − 48rt + 9t2
35. 64r10 − 361s10
3. 16a2 + 56ab + 49b2
37. (19x + 23)(19x − 23)
5. s6 − 18s3 + 81
39. (4v + 13)(4v − 13)
7. s4 + 12s2 t2 + 36t4
41. (13x + 24y)(13x − 24y)
9. (5s + 6t)2
43. (23r + 17s)(23r − 17s)
11. (6v − 5w)2
45. (7r3 + 16t3 )(7r3 − 16t3 )
13. (a2 + 9b2 )2
47. (6u5 + 5w5 )(6u5 − 5w5 )
15. (7s2 − 2t2 )2
49. 2y 3 (6y + 11)(6y − 11)
17. (7b3 − 8)2
51. 4ab(19a + 9b)(19a − 9b)
19. (7r3 + 8)2
53. 4xz(12x + 17z)(12x − 17z)
21. 5st(s − 2t)2
23. 2ac(2a + c)2
25. −3b(4b − 5)2
27. −5u3 (u + 3)2
29. 441c2 − 256
31. 25x2 − 361z 2
33. 9y 8 − 529z 8
55. 4t2 (12t + 1)(12t − 1)
57. (9x2 + 16)(3x + 4)(3x − 4)
59. (9x2 + 4)(3x + 2)(3x − 2)
61. (z + 3)(z − 3)(z + 1)
63. (x + y)(x − y)(x − 2y)
65. (r + 5t)(r − 5t)(r − 3t)
67. (x + 4)(x − 4)(2x + 1)
455
6.5. FACTORING SPECIAL FORMS
69. x = −6, 6, −
7
2
71. x = −8, 8, −5
73. x =
11
12
75. x = −
13 13
,
4 4
5 5
77. x = − ,
3 3
79. x = −
19
16
81. x = 0, −1, 1
83. x = 0, −1/2, 1/2