4 Factoring the Trinomial ax[Sup(2)] + bx + c with a ≠ 1
Tải bản đầy đủ - 0trang
dug84356_ch05a.qxd
348
9/14/10
12:27 PM
Page 348
5-28
Chapter 5 Factoring
E X A M P L E
1
The ac method
Factor each trinomial.
a) 2x2 ϩ 7x ϩ 6
b) 2x2 ϩ x Ϫ 6
c) 10x2 ϩ 13x Ϫ 3
Solution
a) In 2x2 ϩ 7x ϩ 6 we have a ϭ 2, b ϭ 7, and c ϭ 6. So,
ac ϭ 2 и 6 ϭ 12.
Now we need two integers with a product of 12 and a sum of 7. The pairs of
integers with a product of 12 are 1 and 12, 2 and 6, and 3 and 4. Only 3 and 4 have
a sum of 7. Replace 7x by 3x ϩ 4x and factor by grouping:
2x2 ϩ 7x ϩ 6 ϭ 2x2 ϩ 3x ϩ 4x ϩ 6
Replace 7x by 3x ϩ 4x.
ϭ (2x ϩ 3)x ϩ (2x ϩ 3)2 Factor out the common factors.
ϭ (2x ϩ 3)(x ϩ 2)
Factor out 2x ϩ 3.
Check by FOIL.
b) In 2x2 ϩ x Ϫ 6 we have a ϭ 2, b ϭ 1, and c ϭ Ϫ6. So,
ac ϭ 2(Ϫ6) ϭ Ϫ12.
Now we need two integers with a product of Ϫ12 and a sum of 1. We can list the
possible pairs of integers with a product of Ϫ12 as follows:
1 and Ϫ12
Ϫ1 and 12
2 and Ϫ6
3 and Ϫ4
Ϫ2 and 6
Ϫ3 and 4
Only Ϫ3 and 4 have a sum of 1. Replace x by Ϫ3x ϩ 4x and factor by grouping:
2x2 ϩ x Ϫ 6 ϭ 2x2 Ϫ 3x ϩ 4x Ϫ 6
Replace x by Ϫ3x ϩ 4x.
ϭ (2x Ϫ 3)x ϩ (2x Ϫ 3)2 Factor out the common factors.
ϭ (2x Ϫ 3)(x ϩ 2)
Factor out 2x Ϫ 3.
Check by FOIL.
c) Because ac ϭ 10(Ϫ3) ϭ Ϫ30, we need two integers with a product of Ϫ30 and a
sum of 13. The product is negative, so the integers must have opposite signs. We
can list all pairs of factors of Ϫ30 as follows:
1 and Ϫ30
Ϫ1 and 30
2 and Ϫ15
Ϫ2 and 15
3 and Ϫ10
Ϫ3 and 10
5 and Ϫ6
Ϫ5 and 6
The only pair that has a sum of 13 is Ϫ2 and 15:
10x2 ϩ 13x Ϫ 3 ϭ 10x2 Ϫ 2x ϩ 15x Ϫ 3
Replace 13x by Ϫ2x ϩ 15x.
ϭ (5x Ϫ 1)2x ϩ (5x Ϫ 1)3 Factor out the common factors.
ϭ (5x Ϫ 1)(2x ϩ 3)
Factor out 5x Ϫ 1.
Check by FOIL.
Now do Exercises 1–38
dug84356_ch05a.qxd
9/14/10
12:27 PM
Page 349
5-29
E X A M P L E
5.4
2
Factoring the Trinomial ax2 ϩ bx ϩ c with a
1
349
Factoring a trinomial in two variables by the ac method
Factor 8x2 Ϫ 14xy ϩ 3y2
Solution
Since a ϭ 8, b ϭ Ϫ14, and c ϭ 3, we have ac ϭ 24. Two numbers with a product of
24 and a sum of Ϫ14 must both be negative. The possible pairs with a product of 24
follow:
Ϫ1 and Ϫ24
Ϫ3 and Ϫ8
Ϫ2 and Ϫ12
Ϫ4 and Ϫ6
Only Ϫ2 and Ϫ12 have a sum of Ϫ14. Replace Ϫ14xy by Ϫ2xy Ϫ 12xy and factor by
grouping:
8x2 Ϫ 14xy ϩ 3y2 ϭ 8x2 Ϫ 2xy Ϫ 12xy ϩ 3y2
ϭ (4x Ϫ y)2x ϩ (4x Ϫ y)(Ϫ3y)
ϭ (4x Ϫ y)(2x Ϫ 3y)
Check by FOIL.
Now do Exercises 39–44
U2V Trial and Error
After you have gained some experience at factoring by the ac method, you can often
find the factors without going through the steps of grouping. For example, consider
the polynomial
3x2 ϩ 7x Ϫ 6.
The factors of 3x2 can only be 3x and x. The factors of 6 could be 2 and 3 or 1 and 6.
We can list all of the possibilities that give the correct first and last terms, without
regard to the signs:
(3x
3)(x 2)
(3x 2)(x 3)
(3x 6)(x 1)
(3x 1)(x 6)
Because the factors of Ϫ6 have unlike signs, one binomial factor is a sum and the
other binomial is a difference. Now we try some products to see if we get a middle
term of 7x:
(3x ϩ 3)(x Ϫ 2) ϭ 3x2 Ϫ 3x Ϫ 6 Incorrect
(3x Ϫ 3)(x ϩ 2) ϭ 3x2 ϩ 3x Ϫ 6 Incorrect
U Helpful Hint V
If the trinomial has no common factor, then neither binomial factor can
have a common factor.
Actually, there is no need to try (3x 3)(x 2) or (3x 6)(x 1) because each contains
a binomial with a common factor. A common factor in the binomial causes a common
factor in the product. But 3x2 ϩ 7x Ϫ 6 has no common factor. So the factors must
come from either (3x 2)(x 3) or (3x 1)(x 6). So we try again:
(3x ϩ 2)(x Ϫ 3) ϭ 3x2 Ϫ 7x Ϫ 6 Incorrect
(3x Ϫ 2)(x ϩ 3) ϭ 3x2 ϩ 7x Ϫ 6 Correct
Even though there may be many possibilities in some factoring problems, it is
often possible to find the correct factors without writing down every possibility. We
can use a bit of guesswork in factoring trinomials. Try whichever possibility you think
might work. Check it by multiplying. If it is not right, then try again. That is why this
method is called trial and error.
dug84356_ch05a.qxd
350
9/14/10
12:27 PM
Page 350
5-30
Chapter 5 Factoring
E X A M P L E
3
Trial and error
Factor each trinomial using trial and error.
a) 2x2 ϩ 5x Ϫ 3
b) 3x2 Ϫ 11x ϩ 6
Solution
a) Because 2x2 factors only as 2x и x and 3 factors only as 1 и 3, there are only two
possible ways to get the correct first and last terms, without regard to the signs:
U Helpful Hint V
The ac method is more systematic
than trial and error. However, trial
Helpful Hint
U
and error can beVfaster and easier,
especially
if your isfirst
or second
trial
The
ac method
more
systematic
is correct.
than
trial and error. However, trial
and error can be faster and easier,
especially if your first or second trial
is correct.
(2x
1)(x 3)
and
(2x 3)(x 1)
Because the last term of the trinomial is negative, one of the missing signs must
be ϩ, and the other must be Ϫ. The trinomial is factored correctly as
2x2 ϩ 5x Ϫ 3 ϭ (2x Ϫ 1)(x ϩ 3).
Check by using FOIL.
b) There are four possible ways to factor 3x2 Ϫ 11x ϩ 6:
(3x
1)(x 6)
(3x 2)(x 3)
(3x
6)(x 1)
(3x 3)(x 2)
The first binomials of (3x 6)(x 1) and (3x 3)(x 2) have a common factor of 3.
Since there is no common factor in 3x2 Ϫ11x ϩ 6, we can rule out both of these possibilities. Since the last term in 3x2 Ϫ 11x ϩ 6 is positive and the middle term is negative, both signs in the factors must be negative. So the correct factorization is either
(3x Ϫ 1)(x Ϫ 6) or (3x Ϫ 2)(x Ϫ 3). By using FOIL we can verify that (3x Ϫ 2)(x Ϫ 3) ϭ
3x2 Ϫ 11x ϩ 6. So the polynomial is factored correctly as
3x2 Ϫ 11x ϩ 6 ϭ (3x Ϫ 2)(x Ϫ 3).
Now do Exercises 45–64
Factoring by trial and error is not just guessing. In fact, if the trinomial has a positive
leading coefficient, we can determine in advance whether its factors are sums or
differences.
Using Signs in Trial and Error
1. If the signs of the terms of a trinomial are ϩ ϩ ϩ, then both factors are
sums: x2 ϩ 5x ϩ 6 ϭ (x ϩ 2)(x ϩ 3).
2. If the signs are ϩ Ϫ ϩ, then both factors are differences: x2 Ϫ 5x ϩ 6 ϭ
(x Ϫ 2)(x Ϫ 3).
3. If the signs are ϩ ϩ Ϫ or ϩ Ϫ Ϫ, then one factor is a sum and the
other is a difference: x2 ϩ x Ϫ 6 ϭ (x ϩ 3)(x Ϫ 2) and x2 Ϫ x Ϫ 6 ϭ
(x Ϫ 3)(x ϩ 2).
In Example 4 we factor a trinomial that has two variables.
dug84356_ch05a.qxd
9/14/10
12:27 PM
Page 351
5-31
E X A M P L E
5.4
4
Factoring the Trinomial ax2 ϩ bx ϩ c with a
1
351
Factoring a trinomial with two variables by trial and error
Factor 6x2 Ϫ 7xy ϩ 2y2.
Solution
We list the possible ways to factor the trinomial:
(3x
2y)(2x y)
(3x y)(2x 2y)
(6x 2y)(x y)
(6x y)(x 2y)
Note that there is a common factor 2 in (2x 2y) and in (6x 2y). Since there is no common factor of 2 in the original trinomial, the second and third possibilities will not work.
Because the last term of the trinomial is positive and the middle term is negative, both factors must contain subtraction symbols. Only the first possibility will give a middle term of
Ϫ7xy when subtraction symbols are used in both factors. So,
6x2 Ϫ 7xy ϩ 2y2 ϭ (3x Ϫ 2y)(2x Ϫ y).
Now do Exercises 65–74
U3V Factoring Completely
You can use the latest factoring technique along with the techniques that you learned
earlier to factor polynomials completely. Remember always to first factor out the
greatest common factor (if it is not 1).
E X A M P L E
5
Factoring completely
Factor each polynomial completely.
a) 4x3 ϩ 14x2 ϩ 6x
b) 12x2y ϩ 6xy ϩ 6y
Solution
a) 4x3 ϩ 14x2 ϩ 6x ϭ 2x(2x2 ϩ 7x ϩ 3)
ϭ 2x(2x ϩ 1)(x ϩ 3)
Factor out the GCF, 2x.
Factor 2x2 ϩ 7x ϩ 3.
Check by multiplying.
b) 12x 2y ϩ 6xy ϩ 6y ϭ 6y(2x2 ϩ x ϩ 1) Factor out the GCF, 6y.
To factor 2x 2 ϩ x ϩ 1 by the ac method, we need two numbers with a product of 2
and a sum of 1. Because there are no such numbers, 2x2 ϩ x ϩ 1 is prime and the
factorization is complete.
Now do Exercises 75–84
Our first step in factoring is to factor out the greatest common factor (if it is not 1).
If the first term of a polynomial has a negative coefficient, then it is better to factor out
the opposite of the GCF so that the resulting polynomial will have a positive leading
coefficient.
dug84356_ch05a.qxd 9/17/10 7:56 PM Page 352
352
5-32
Chapter 5 Factoring
E X A M P L E
6
Factoring out the opposite of the GCF
Factor each polynomial completely.
a) Ϫ18x3 ϩ 51x2 Ϫ 15x
b) Ϫ3a2 ϩ 2a ϩ 21
Solution
a) The GCF is 3x. Because the first term has a negative coefficient, we factor
out Ϫ3x:
Ϫ18x3 ϩ 51x2 Ϫ 15x ϭ Ϫ3x(6x2 Ϫ 17x ϩ 5)
Factor out Ϫ3x.
ϭ Ϫ3x(3x Ϫ 1)(2x Ϫ 5) Factor 6x2 Ϫ 17x ϩ 5.
b) The GCF for Ϫ3a2 ϩ 2a ϩ 21 is 1. Because the first term has a negative coefficient, factor out Ϫ1:
Ϫ3a2 ϩ 2a ϩ 21 ϭ Ϫ1(3a2 Ϫ 2a Ϫ 21) Factor out Ϫ1.
ϭ Ϫ1(3a ϩ 7)(a Ϫ 3)
Factor 3a2 Ϫ 2a Ϫ 21.
Now do Exercises 85–100
Warm-Ups
▼
Fill in the blank.
5.4
1. If there are no two integers that have a
of ac
and a
of b, then ax2 ϩ bx ϩ c is prime.
2. In the
method we make educated
guesses at the factors and then check by FOIL.
True or false?
3.
4.
5.
6.
7.
8.
2x2 ϩ 3x ϩ 1 ϭ (2x ϩ 1)(x ϩ 1)
2x2 ϩ 5x ϩ 3 ϭ (2x ϩ 1)(x ϩ 3)
3x2 ϩ 10x ϩ 3 ϭ (3x ϩ 1)(x ϩ 3)
2x2 Ϫ 7x Ϫ 9 ϭ (2x Ϫ 9)(x ϩ 1)
2x2 Ϫ 16x Ϫ 9 ϭ (2x Ϫ 9)(2x ϩ 1)
12x2 Ϫ 13x ϩ 3 ϭ (3x Ϫ 1)(4x Ϫ 3)
Exercises
U Study Tips V
• Pay particular attention to the examples that your instructor works in class or presents to you online.
• The examples and homework assignments should give you a good idea of what your instructor expects from you.
U1V The ac Method
Find the following. See Example 1.
1. Two integers that have a product of 12 and a sum of 7
2. Two integers that have a product of 20 and a sum of 12
3. Two integers that have a product of 30 and a sum of Ϫ17
4. Two integers that have a product of 36 and a sum of Ϫ20
5. Two integers that have a product of Ϫ12 and a sum of Ϫ4
dug84356_ch05a.qxd
9/14/10
12:27 PM
Page 353
5-33
5.4
6. Two integers that have a product of Ϫ8 and a sum
of 7
Each of the following trinomials is in the form ax2 ϩ bx ϩ c.
For each trinomial, find two integers that have a product of ac
and a sum of b. Do not factor the trinomials. See Example 1.
Factoring the Trinomial ax2 ϩ bx ϩ c with a
1
47. 6x2 ϩ 5x ϩ 1
48. 15y2 ϩ 8y ϩ 1
49. 5a2 ϩ 11a ϩ 2
50. 3y2 ϩ 10y ϩ 7
51. 4w2 ϩ 8w ϩ 3
52. 6z2 ϩ 13z ϩ 5
7. 6x2 ϩ 7x ϩ 2
8. 5x2 ϩ 17x ϩ 6
53. 15x2 Ϫ x Ϫ 2
54. 15x2 ϩ 13x Ϫ 2
9. 6y2 Ϫ 11y ϩ 3
10. 6z2 Ϫ 19z ϩ 10
55. 8x2 Ϫ 6x ϩ 1
56. 8x2 Ϫ 22x ϩ 5
11. 12w2 ϩ w Ϫ 1
12. 15t2 Ϫ 17t Ϫ 4
57. 15x2 Ϫ 31x ϩ 2
58. 15x2 ϩ 31x ϩ 2
353
Factor each trinomial using the ac method. See Example 1.
See the Strategy for Factoring ax2 ϩ bx ϩ c by the ac Method
box on page 347.
13. 2x2 ϩ 3x ϩ 1
14. 2x2 ϩ 11x ϩ 5
59. 4x2 Ϫ 4x ϩ 3
60. 4x2 ϩ 12x Ϫ 5
61. 2x2 ϩ 18x Ϫ 90
62. 3x2 ϩ 11x ϩ 10
63. 3x2 ϩ x Ϫ 10
64. 3x2 Ϫ 17x ϩ 10
15. 2x2 ϩ 9x ϩ 4
16. 2h2 ϩ 7h ϩ 3
65. 10x2 Ϫ 3xy Ϫ y2
66. 8x2 Ϫ 2xy Ϫ y2
17. 3t 2 ϩ 7t ϩ 2
18. 3t2 ϩ 8t ϩ 5
67. 42a2 Ϫ 13ab ϩ b2
68. 10a2 Ϫ 27ab ϩ 5b2
19. 2x2 ϩ 5x Ϫ 3
20. 3x2 Ϫ x Ϫ 2
21. 6x2 ϩ 7x Ϫ 3
22. 21x2 ϩ 2x Ϫ 3
23. 3x2 Ϫ 5x ϩ 4
25. 2x2 Ϫ 7x ϩ 6
24. 6x2 Ϫ 5x ϩ 3
26. 3a2 Ϫ 14a ϩ 15
27. 5b2 Ϫ 13b ϩ 6
28. 7y2 ϩ 16y Ϫ 15
29. 4y2 Ϫ 11y Ϫ 3
30. 35x2 Ϫ 2x Ϫ 1
31. 3x ϩ 2x ϩ 1
33. 8x2 Ϫ 2x Ϫ 1
32. 6x Ϫ 4x Ϫ 5
34. 8x2 Ϫ 10x Ϫ 3
Factor each polynomial completely. See Examples 5 and 6.
75. 81w3 Ϫ w
76. 81w3 Ϫ w2
35. 9t2 Ϫ 9t ϩ 2
36. 9t2 ϩ 5t Ϫ 4
77. 4w2 ϩ 2w Ϫ 30
78. 2x2 Ϫ 28x ϩ 98
37. 15x2 ϩ 13x ϩ 2
38. 15x2 Ϫ 7x Ϫ 2
79. 27 ϩ 12x2 ϩ 36x
80. 24y ϩ 12y2 ϩ 12
81. 6w2 Ϫ 11w Ϫ 35
82. 8y2 Ϫ 14y Ϫ 15
2
2
Use the ac method to factor each trinomial. See Example 2.
Complete the factoring.
69.
70.
71.
72.
73.
74.
3x2 ϩ 7x ϩ 2 ϭ (x ϩ 2)(
2x2 Ϫ x Ϫ 15 ϭ (x Ϫ 3)(
5x2 ϩ 11x ϩ 2 ϭ (5x ϩ 1)(
4x2 Ϫ 19x Ϫ 5 ϭ (4x ϩ 1)(
6a2 Ϫ 17a ϩ 5 ϭ (3a Ϫ 1)(
4b2 Ϫ 16b ϩ 15 ϭ (2b Ϫ 5)(
)
)
)
)
)
)
U3V Factoring Completely
39. 4a2 ϩ 16ab ϩ 15b2
40. 10x2 ϩ 17xy ϩ 3y2
83. 3x2z Ϫ 3zx Ϫ 18z
84. a2b ϩ 2ab Ϫ 15b
41. 6m2 Ϫ 7mn Ϫ 5n2
42. 3a2 ϩ 2ab Ϫ 21b2
85. 9x3 Ϫ 21x2 ϩ 18x
86. Ϫ8x3 ϩ 4x2 Ϫ 2x
43. 3x2 Ϫ 8xy ϩ 5y2
44. 3m2 Ϫ 13mn ϩ 12n2
87. a2 ϩ 2ab Ϫ 15b2
88. a2b2 Ϫ 2a2b Ϫ 15a2
89. 2x2y2 ϩ xy2 ϩ 3y2
90. 18x2 Ϫ 6x ϩ 6
91. Ϫ6t3 Ϫ t2 ϩ 2t
92. Ϫ36t2 Ϫ 6t ϩ 12
93. 12t4 Ϫ 2t3 Ϫ 4t 2
94. 12t3 ϩ 14t2 ϩ 4t
U2V Trial and Error
Factor each trinomial using trial and error. See Examples 3 and 4.
45. 5a2 ϩ 6a ϩ 1
46. 7b2 ϩ 8b ϩ 1
dug84356_ch05a.qxd
354
95.
96.
97.
98.
99.
100.
9/14/10
12:27 PM
Page 354
5-34
Chapter 5 Factoring
4x2y Ϫ 8xy2 ϩ 3y3
9x2 ϩ 24xy Ϫ 9y2
Ϫ4w2 ϩ 7w Ϫ 3
Ϫ30w2 ϩ w ϩ 1
Ϫ12a3 ϩ 22a 2b Ϫ 6ab2
Ϫ36a2b ϩ 21ab2 Ϫ 3b3
a) Rewrite the formula by factoring the right-hand side
completely.
b) Use the factored version of the formula to find N(3).
c) Use the accompanying graph to estimate the time at
which the workers are most efficient.
d) Use the accompanying graph to estimate the
maximum number of components assembled per
hour during an 8-hour shift.
Applications
Solve each problem.
101. Height of a ball. If a ball is thrown straight upward at
40 feet per second from a rooftop 24 feet above the
ground, then its height in feet above the ground t seconds
after it is thrown is given by
h(t) ϭ Ϫ16t2 ϩ 40t ϩ 24.
a) Find h(0), h(1), h(2), and h(3).
b) Rewrite the formula with the polynomial factored
completely.
c) Find h(3) using the result of part (b).
Getting More Involved
103. Exploration
Find all positive and negative integers b for which
each polynomial can be factored.
a) x2 ϩ bx ϩ 3
c) 2x2 ϩ bx Ϫ 15
b) 3x2 ϩ bx ϩ 5
104. Exploration
40 ft/sec
Find two integers c (positive or negative) for which
each polynomial can be factored. Many answers are
possible.
a) x 2 ϩ x ϩ c
b) x2 Ϫ 2x ϩ c
c) 2x 2 Ϫ 3x ϩ c
h(t) ϭ Ϫ16 t 2 ϩ 40t ϩ 24
105. Cooperative learning
Working in groups, cut two large squares, three
rectangles, and one small square out of paper that are
exactly the same size as shown in the accompanying
figure. Then try to place the six figures next to one
another so that they form a large rectangle. Do not
overlap the pieces or leave any gaps. Explain how
factoring 2x2 ϩ 3x ϩ 1 can help you solve this puzzle.
Figure for Exercise 101
102. Worker efficiency. In a study of worker efficiency at Wong
Laboratories it was found that the number of components
assembled per hour by the average worker t hours after
starting work could be modeled by the formula
N(t) ϭ Ϫ3t3 ϩ 23t2 ϩ 8t.
x
x
Number of components
300
1
1
1
1
1
x
x
x
x
x
200
Figure for Exercise 105
100
0
106. Cooperative learning
0 1 2 3 4 5 6 7 8
Time (hours)
Figure for Exercise 102
Working in groups, cut four squares and eight
rectangles out of paper as in Exercise 105 to illustrate the
trinomial 4x2 ϩ 7x ϩ 3. Select one group to demonstrate
how to arrange the 12 pieces to form a large rectangle.
Have another group explain how factoring the trinomial
can help you solve this puzzle.
dug84356_ch05b.qxd
9/14/10
12:31 PM
Page 355
5-35
5.5
5.5
In This Section
U1V Factoring a Difference or Sum
of Two Cubes
2
U V Factoring a Difference of Two
Fourth Powers
3
U V The Factoring Strategy
Difference and Sum of Cubes and a Strategy
355
Difference and Sum of Cubes and a Strategy
In Sections 5.1 to 5.4, we established the general idea of factoring and some special
cases. In this section we will see two more special cases. We will then summarize all
of the factoring that we have done with a factoring strategy.
U1V Factoring a Difference or Sum of Two Cubes
We can use division to discover that a Ϫ b is a factor of a3 Ϫ b3 (a difference of two
cubes) and a ϩ b is a factor of a3 ϩ b3 (a sum of two cubes):
a2 ϩ ab ϩ b2
3 ෆෆෆෆ
ෆෆ
a Ϫ bͤෆaෆ
ϩ 0a2bෆ
ϩෆ0aෆbෆ2ෆϪ
bෆ3
3
2
a Ϫ ab
a2b ϩ 0ab2
a2b Ϫ ab2
ab2 Ϫ b3
ab2 Ϫ b3
0
a2 Ϫ ab ϩ b2
2 ෆෆෆෆ2ෆෆෆෆ
ͤ
a ϩ b ෆa3ෆෆ
ϩෆ0ෆaෆ
b ϩ 0ab ϩ b3
3
2
a ϩ ab
Ϫa2b ϩ 0ab2
Ϫa2b Ϫ ab2
ab2 ϩ b3
ab2 ϩ b3
0
In each division the remainder is 0. So in each case the dividend is equal to the divisor times the quotient. These results give us two new factoring rules.
Factoring a Difference or Sum of Two Cubes
a3 Ϫ b3 ϭ (a Ϫ b)(a2 ϩ ab ϩ b2)
a3 ϩ b3 ϭ (a ϩ b)(a2 Ϫ ab ϩ b2)
Use the following strategy to factor a difference or sum of two cubes.
Strategy for Factoring a3 ؊ b3 or a3 ؉ b3
1. The first factor is the original polynomial without the exponents, and the
middle term in the second factor has the opposite sign from the first factor:
a3 – b3 ϭ (a – b)(a2 ϩ ab ϩ b2)
↑
↑
opposite signs
a3 ϩ b3 ϭ (a ϩ b)(a2 – ab ϩ b2)
↑
↑
opposite signs
2. Recall the two perfect square trinomials a ϩ 2ab ϩ b2 and a2 – 2ab ϩ b2.
2
The second factor is almost a perfect square trinomial. Just delete the 2.
It is helpful also to compare the differences and sums of squares and cubes:
a2 Ϫ b2 ϭ (a Ϫ b)(a ϩ b)
a3 Ϫ b3 ϭ (a Ϫ b)(a2 ϩ ab ϩ b2)
a2 ϩ b2 Prime
a3 ϩ b3 ϭ (a ϩ b)(a2 Ϫ ab ϩ b2)
The factors a2 ϩ ab ϩ b2 and a2 Ϫ ab ϩ b2 are prime. They can’t be factored. The
perfect square trinomials a2 ϩ 2ab ϩ b2 and a2 Ϫ 2ab ϩ b2, which are almost the
same, are not prime. They can be factored:
a2 ϩ 2ab ϩ b2 ϭ (a ϩ b)2
and
a2 Ϫ 2ab ϩ b2 ϭ (a Ϫ b)2.
dug84356_ch05b.qxd
356
9/14/10
12:31 PM
Page 356
5-36
Chapter 5 Factoring
E X A M P L E
1
Factoring a difference or sum of two cubes
Factor each polynomial.
a) w3 Ϫ 8
b) x3 ϩ 1
c) 8y3 Ϫ 27
Solution
a) Because 8 ϭ 23, w3 Ϫ 8 is a difference of two cubes. To factor w3 Ϫ 8, let
a ϭ w and b ϭ 2 in the formula a3 Ϫ b3 ϭ (a Ϫ b)(a2 ϩ ab ϩ b2):
w3 Ϫ 8 ϭ (w Ϫ 2)(w2 ϩ 2w ϩ 4)
b) Because 1 ϭ 13, the binomial x3 ϩ 1 is a sum of two cubes. Let a ϭ x and b ϭ 1 in
the formula a3 ϩ b3 ϭ (a ϩ b)(a2 Ϫ ab ϩ b2):
x3 ϩ 1 ϭ (x ϩ 1)(x2 Ϫ x ϩ 1)
c) 8y3 Ϫ 27 ϭ (2y)3 Ϫ 33
This is a difference of two cubes.
ϭ (2y Ϫ 3)(4y ϩ 6y ϩ 9) Let a ϭ 2y and b ϭ 3 in the formula.
2
Now do Exercises 1–16
In Example 1, we used the first three perfect cubes, 1, 8, and 27. You should verify
that 1, 8, 27, 64, 125, 216, 343, 512, 729, and 1000 are the first 10 perfect cubes.
CAUTION The polynomial (a Ϫ b)3 is not equivalent to a3 Ϫ b3 because if a ϭ 2
and b ϭ 1, then
(a Ϫ b)3 ϭ (2 Ϫ 1)3 ϭ 13 ϭ 1
and
a3 Ϫ b3 ϭ 23 Ϫ 13 ϭ 8 Ϫ 1 ϭ 7.
Likewise, (a ϩ b)3 is not equivalent to a3 ϩ b3.
U2V Factoring a Difference of Two Fourth Powers
A difference of two fourth powers of the form a4 Ϫ b4 is also a difference of two
squares, (a2)2 Ϫ (b2)2. It can be factored by the rule for factoring a difference of two
squares:
Write as a difference of two squares.
a4 Ϫ b4 ϭ (a2)2 Ϫ (b2)2
2
2
2
2
ϭ (a Ϫ b )(a ϩ b )
Difference of two squares
ϭ (a Ϫ b)(a ϩ b)(a2 ϩ b2) Factor completely.
Note that the sum of two squares a2 ϩ b2 is prime and cannot be factored.
E X A M P L E
2
Factoring a difference of two fourth powers
Factor each polynomial completely.
a) x4 Ϫ 16
b) 81m4 Ϫ n4
dug84356_ch05b.qxd
9/14/10
12:31 PM
Page 357
5-37
5.5
Difference and Sum of Cubes and a Strategy
357
Solution
a) x4 Ϫ 16 ϭ (x2)2 Ϫ 42
Write as a difference of two squares.
ϭ (x Ϫ 4)(x ϩ 4)
2
2
Difference of two squares
ϭ (x Ϫ 2)(x ϩ 2)(x ϩ 4)
2
Factor completely.
b) 81m4 Ϫ n4 ϭ (9m2)2 Ϫ (n2)2
ϭ (9m Ϫ n
2
Write as a difference of two squares.
)(9m
2
2
ϩn
2
)
Factor.
ϭ (3m Ϫ n)(3m ϩ n)(9m2 ϩ n2)
Factor completely.
Now do Exercises 17–24
CAUTION A difference of two squares or cubes can be factored, and a sum of two
cubes can be factored. But the sums of two squares x2 ϩ 4 and 9m2 ϩ n2
in Example 2 are prime.
U3V The Factoring Strategy
The following is a summary of the ideas that we use to factor a polynomial completely.
Strategy for Factoring Polynomials Completely
1. Factor out the GCF (with a negative coefficient if necessary).
2. When factoring a binomial, check to see whether it is a difference of two
3.
4.
5.
6.
squares, a difference of two cubes, or a sum of two cubes. A sum of two
squares does not factor.
When factoring a trinomial, check to see whether it is a perfect square trinomial.
If the polynomial has four terms, try factoring by grouping.
When factoring a trinomial that is not a perfect square, use the ac method or
the trial-and-error method.
Check to see whether any of the factors can be factored again.
We will use the factoring strategy in Example 3.
E X A M P L E
3
Factoring polynomials
Factor each polynomial completely.
a) 2a2b Ϫ 24ab ϩ 72b
b) 3x3 ϩ 6x2 Ϫ 75x Ϫ 150
c) Ϫ3x4 Ϫ 15x3 ϩ 72x2
d) 60y3 Ϫ 85y2 Ϫ 25y
Solution
a) 2a2b Ϫ 24ab ϩ 72b ϭ 2b(a2 Ϫ 12a ϩ 36)
ϭ 2b(a Ϫ 6)2
First factor out the GCF, 2b.
Factor the perfect square
trinomial.
b) 3x3 ϩ 6x2 Ϫ 75x Ϫ 150 ϭ 3[x3 ϩ 2x 2 Ϫ 25x Ϫ 50] Factor out the GCF, 3.
ϭ 3[x 2(x ϩ 2) Ϫ 25(x ϩ 2)] Factor out common factors.
ϭ 3(x 2 Ϫ 25)(x ϩ 2)
Factor by grouping.
ϭ 3(x ϩ 5)(x Ϫ 5)(x ϩ 2)
Factor the difference
of two squares.
dug84356_ch05b.qxd 9/17/10 7:57 PM Page 358
358
5-38
Chapter 5 Factoring
c) Factor out Ϫ3x2 to get Ϫ3x4 Ϫ 15x3 ϩ 72x2 ϭ Ϫ3x2(x2 ϩ 5x Ϫ 24). To factor the
trinomial, find two numbers with a product of Ϫ24 and a sum of 5. For a product
of 24 we have 1 и 24, 2 и 12, 3 и 8, and 4 и 6. To get a sum of 5 and a product of
Ϫ24 choose 8 and Ϫ3:
Ϫ3x4 Ϫ 15x3 ϩ 72x2 ϭ Ϫ3x2(x2 ϩ 5x Ϫ 24)
ϭ Ϫ3x2(x Ϫ 3)(x ϩ 8)
d) Factor out 5y to get 60y3 Ϫ 85y2 Ϫ 25y ϭ 5y(12y2 Ϫ 17y Ϫ 5). By the ac method
we need two numbers that have a product of Ϫ60 (ac) and a sum of Ϫ17. The
numbers are Ϫ20 and 3. Now factor by grouping:
60y3 Ϫ 85y2 Ϫ 25y ϭ 5y(12y2 Ϫ 17y Ϫ 5)
ϭ 5y(12y2 Ϫ 20y ϩ 3y Ϫ 5)
ϭ 5y[4y(3y Ϫ 5) ϩ 1(3y Ϫ 5)]
ϭ 5y(3y Ϫ 5)(4y ϩ 1)
Factor out 5y.
Ϫ17y ϭ Ϫ20y ϩ 3y
Factor by grouping.
Factor out 3y Ϫ 5.
Now do Exercises 25–92
Warm-Ups
▼
Fill in the blank.
5.5
1. If there is no
, then the dividend is the divisor
times the quotient.
2. The binomial a3 ϩ b3 is a
of two cubes.
3
3
3. The binomial a Ϫ b is a
of two cubes.
4. If a3 Ϫ b3 is divided by a Ϫ b, then the remainder
is
.
True or false?
5. For any real number x, x2 Ϫ 4 ϭ (x Ϫ 2)2.
6. The trinomial 4x2 ϩ 6x ϩ 9 is a perfect square
trinomial.
7. The binomial 4y2 ϩ 25 is prime.
8. If the GCF is not 1, then you should factor it out
first.
9. You can factor y2 Ϫ 5y Ϫ my ϩ 5m by grouping.
10. You can factor x2 ϩ ax Ϫ 3x ϩ 3a by grouping.
Exercises
U Study Tips V
• If you have a choice, sit at the front of the class. It is easier to stay alert when you are at the front.
• If you miss what is going on in class, you miss what your instructor feels is important and most likely to appear on tests and quizzes.
U1V Factoring a Difference or Sum of Two Cubes
Factor each difference or sum of cubes. See Example 1.
1. m3 Ϫ 1
2. z3 Ϫ 27
3.
4.
5.
6.
x3 ϩ 8
y3 ϩ 27
a3 ϩ 125
b3 Ϫ 216