2 Factoring: Common Factor and Difference of Squares
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CHAPTER 6 Factoring and Fractions
E X A M P L E 1 Factoring completely
When we factor the expression 12x + 6x2 as
12x + 6x2 = 216x + 3x2 2
we see that it has not been factored completely. The factor 6x + 3x2 is not prime,
because it may be factored as
6x + 3x2 = 3x12 + x2
Therefore, the expression 12x + 6x2 is factored completely as
12x + 6x2 = 6x12 + x2
LEARNING TIP
The ability to factor algebraic expressions depends heavily on the proper
recognition of the special products.
■ For reference, Eq. (6.1) is
a1 x + y 2 = ax + ay.
The factors x and 12 + x2 are prime. We can factor the numerical coefficient, 6, as
2132, but it is standard not to write numerical coefficients in factored form.
■
To factor expressions easily, we must know how to do algebraic multiplication and
really know the special products of the previous section. The special products also give
us a way of checking answers and deciding whether a given factor is prime.
COMMON MONOMIAL FACTORS
Often an expression contains a monomial that is common to each term of the expression.
Therefore, the first step in factoring any expression should be to factor out any common
monomial factor that may exist. To do this, we note the common factor by inspection
and then use the reverse of the distributive law, Eq. (6.1), to show the factored form. The
following examples illustrate factoring a common monomial factor out of an
expression.
E X A M P L E 2 Common monomial factor
In factoring 6x - 2y, we note each term contains a factor of 2:
6x - 2y = 213x2 - 2y = 213x - y2
LEARNING TIP
Usually, the factorization of common
monomial factors is done by inspection. However, once the common factor is found, the other factor can also
be determined by dividing the original
expression by the common factor. For
example, 2abc + 4ab has a common
factor of 2ab in each term, so
2abc + 4ab = 2aba
2abc
4ab
+
b
2ab
2ab
2abc + 4ab = 2ab(c + 2)
LEARNING TIP
You can check your factoring result by
multiplication.
Practice Exercises
Factor: 1. 3cx3 - 9cx
2. 9cx3 - 3cx
Here, 2 is the common monomial factor, and 213x - y2 is the required factored form
of 6x - 2y. Once the common factor has been identified, it is not actually necessary to
write a term like 6x as 213x2. The result can be written directly.
In this case,
213x - y2 = 6x - 2y
Since the result of the multiplication gives the original expression, the factored form is
correct.
■
The next example illustrates the case where the common factor is the same as one of
the terms, and special care must be taken to complete the factoring correctly.
E X A M P L E 3 Common factor same as term
Factor 4ax2 + 2ax.
The numerical factor 2 and the literal factors a and x are common to each term.
Therefore, the common monomial factor of 4ax2 + 2ax is 2ax. This means that
4ax2 + 2ax = 2ax12x2 + 2ax112 = 2ax12x + 12
Note the presence of the 1 in the factored form. When we divide 4ax2 + 2ax by 2ax,
we get
4ax2 + 2ax
4ax2
2ax
=
+
2ax
2ax
2ax
= 2x + 1
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6.2 Factoring: Common Factor and Difference of Squares
187
noting very carefully that 2ax divided by 2ax is 1 and does not simply cancel out leaving nothing. Without the 1, when the factored form is multiplied out, we would not
obtain the original expression.
■
COMMON ERROR
When the common factor is the same as one of the terms in the expression, remember
that the term divides to 1, so do not omit the 1 and cancel the term out to nothing!
5r t + 5r = 5r (t + 1)
Not
5r t + 5r ≠ 5r (t + 0)
E X A M P L E 4 Common factor by inspection
Factor 6a5x2 - 9a3x3 + 3a3x2.
After inspecting each term, we determine that each contains a factor of 3, a3, and x2.
Thus, the common monomial factor is 3a3x2. This means that
6a5x2 - 9a3x3 + 3a3x2 = 3a3x2 12a2 - 3x + 12
■
In these examples, note that factoring an expression does not actually change the
expression, although it does change the form of the expression. In equating the expression to its factored form, we write an identity.
It is often necessary to use factoring when solving an equation. This is illustrated in
the following example.
E X A M P L E 5 Using factoring in solving an equation
An equation used in the analysis of FM reception is RF = a 12RA + RF 2. Solve for RF.
The steps in the solution are as follows:
RF = a 12RA + RF 2
RF = 2aRA + aRF
■ FM radio was developed in the early 1930s.
RF - aRF = 2aRA
RF 11 - a 2 = 2aRA
RF =
2aRA
1 - a
original equation
use distributive law
subtract aRF from both sides
factor out RF on left
divide both sides by 1 - a
We see that we collected both terms containing RF on the left so that we could factor
and thereby solve for RF.
■
■ For reference, Eq. (6.2) is
1 x + y 2 1 x - y 2 = x 2 - y 2.
LEARNING TIP
Usually, in factoring an expression of
this type, where it is very clear what
numbers are squared, we do not actually write out the middle step as shown.
However, if in doubt, write it out.
FACTORING THE DIFFERENCE OF TWO SQUARES
In Eq. (6.2), we see that the product of the sum and the difference of two numbers results
in the difference between the squares of two numbers. Therefore, factoring the difference
of two squares gives factors that are the sum and the difference of the numbers.
E X A M P L E 6 Factoring the difference of two squares
In factoring x2 - 16, note that x2 is the square of x and 16 is the square of 4. Therefore,
squares
x2 - 16 = x2 - 42 = 1x + 42 1x - 42
difference
sum
difference
■
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CHAPTER 6 Factoring and Fractions
E X A M P L E 7 Factoring the difference of two squares where a factor
is itself a composite factor
2
(a) Since 4x is the square of 2x and 9 is the square of 3, we may factor 4x2 - 9 as
4x2 - 9 = 12x2 2 - 32 = 12x + 32 12x - 32
(b) In the same way,
1y - 32 2 - 16x4 = 3 1y - 32 + 4x2 4 3 1 y - 32 - 4x2 4
= 1y - 3 + 4x2 2 1y - 3 - 4x2 2
where we note that 16x4 = 14x2 2 2.
■
COMPLETE FACTORING
As noted before, a common monomial factor should be factored out first. However,
we must be careful to see if the other factor can itself be factored. Be sure to include
only prime factors in the final result.
COMMON ERROR
Always factor your equations completely. It is a common error to stop before the expression involves only prime factors.
For example, 4a2 - a2x 2 can be factored to a2(4 - x 2), and it is a common error to
think you are now finished. However, the second factor is a difference of squares. The
complete factorization is
4a2 - a2x 2 = a2(4 - x 2)
= a2(2 + x)(2 - x)
E X A M P L E 8 Complete factoring
(a) In factoring 20x2 - 45, note a common factor of 5 in each term. Therefore,
20x2 - 45 = 514x2 - 92. However, the factor 4x2 - 9 itself is the difference of
squares. Therefore, 20x2 - 45 is completely factored as
common factor
Practice Exercises
Factor: 3. 9c2 - 64
4. 18c2 - 128
2
2
20x - 45 = 514x - 92 = 512x + 32 12x - 32
difference of squares
4
4
(b) In factoring x - y , note that we have the difference of two squares. Therefore,
x4 - y4 = 1x2 + y2 2 1x2 - y2 2. However, the factor x2 - y2 is also the difference of squares. This means that
x4 - y4 = 1x2 + y2 2 1x2 - y2 2 = 1x2 + y2 2 1x + y2 1x - y2
COMMON ERROR
■
The factor x 2 + y 2 is prime. It is not equal to 1x + y 2 2. (See Example 2 of Section 6.1.)
FACTORING BY GROUPING
The terms in a polynomial can sometimes be grouped such that the polynomial can then
be factored by the methods of this section. The following example illustrates this
method of factoring by grouping.
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6.2 Factoring: Common Factor and Difference of Squares
189
E X A M P L E 9 Factoring by grouping
Factor 2x - 2y + ax - ay.
We see that there is no common factor to all four terms, but that each of the first two
terms contains a factor of 2, and each of the third and fourth terms contains a factor of a.
Grouping terms this way and then factoring each group, we have
2x - 2y + ax - ay = 12x - 2y2 + 1ax - ay2
= 21x - y2 + a1x - y2
= 1x - y2 12 + a2
now note the common factor of 1x - y2
■
The general method of factoring by grouping can be used with several types of
groupings. We will discuss another type in the following section.
E XE RC IS ES 6 .2
In Exercises 1–4, make the given changes in the indicated examples of
this section and then solve the indicated problems.
1. In Example 3, change the + sign to - and then factor.
2. In Example 3, set the given expression equal to B and then solve for a.
3. In Example 8(a), change the coefficient of the first term from 20
to 5 and then factor.
4. In Example 9, change both - signs to + and then factor.
5. 6x + 6y
6. 3a - 3b
2
8. 2x + 2
2
9. 3x - 9x
10. 4s2 + 20s
2
2
13. 288n2 + 24n
16. 23a - 46b + 69c
14. 90p3 - 15p2
2
19. 12pq - 8pq - 28pq
2
2
3
21. 2a - 2b + 4c - 6d
2
20. 27a b - 24ab - 9a
22. 5a + 10ax - 5ay - 20az
23. x - 4
24. r - 25
25. 100 - 9A
26. 49 - Z 4
27. 36a4 + 1
28. 324z2 - 4
2
29. 162s - 50t
2
2 2
2
30. 36s - 121t
2
32. 36a b + 169c
2
35. 2x - 8
38. 28x2 - 700y2
41. x4 - 16
2
2
In Exercises 63–70, factor the expressions completely. In Exercises 69
and 70, it is necessary to set up the proper expression. Each
expression comes from the technical area indicated.
66. PbL2 - Pb3
67. rR2 - r 3
2
34. 1a - b2 - 1
36. 5a - 125
37. 300x - 2700z
39. 21I - 32 2 - 8
40. a1x + 222 - ay2
42. y4 - 81
43. x8 - 1
2
(container surface area)
3
(machine design)
(business)
(architecture)
(pipeline flow)
68. p1R2 - p1r 2 - p2R2 + p2r 2
4
33. 1x + y2 - 9
2
72 - 52
61. Factor n2 + n, and then explain why it represents a positive even
integer if n is a positive integer.
65. Rv + Rv + Rv
31. 144n - 169p
2
59 - 57
In Exercises 61 and 62, give the required explanations.
2
2
2
60.
64. 4d 2D2 - 4d 3D - d 4
2
2
89 - 88
7
63. 2prh + 2pr 2
18. 4pq - 14q2 - 16pq2
17. 3ab2 - 6ab + 12ab3
2
7. 5a - 5
12. 5a - 20ax
15. 2x + 4y - 8z
59.
62. Factor n3 - n, and then explain why it represents a multiple of 6
if n is an integer greater than 1.
In Exercises 5–44, factor the given expressions completely.
11. 7b h - 28b
In Exercises 59 and 60, evaluate the given expressions by using
factoring. The results may be checked with a calculator.
2
44. 2x4 - 8y4
In Exercises 45–50, solve for the indicated letter.
45. 2a - b = ab + 3, for a
46. n1x + 12 = 5 - x, for x
47. 3 - 2s = 213 - st2, for s
48. k12 - y2 = y12k - 12, for y
(fluid flow)
69. As large as possible a square is cut from a circular metal plate of
radius r. Express in factored form the area of the metal pieces that
are left.
70. A pipe of outside diameter d is inserted into a pipe of inside radius r. Express in factored form the cross-sectional area within
the larger pipe that is outside the smaller pipe.
In Exercises 71–76, solve for the indicated letter. Each equation
comes from the technical area indicated.
71. i1R1 = 1i2 - i1 2R2, for i1
(electricity: ammeter)
49. 1x + 2k2 1x - 22 = x + 3x - 4k, for k
72. nV + n1v = n1V, for n1
(acoustics)
73. 3BY + 5Y = 9BS, for B
(physics: elasticity)
In Exercises 51–58, factor the given expressions by grouping as
illustrated in Example 9.
75. ER = AtT0 - AtT1, for t
2
50. 12x - 3k2 1x + 12 = 2x2 - x - 3, for k
51. 3x - 3y + bx - by
52. am + an + cn + cm
53. a2 + ax - ab - bx
54. 2y - y2 - 6y4 + 12y3
55. x3 + 3x2 - 4x - 12
56. S3 - 5S2 - S + 5
2
2
57. x - y + x - y
2
2
58. 4p - q + 2p + q
74. Sq + Sp = Spq + p, for q
76. R = kT 42 - kT 41, for k
(computer design)
(energy conservation)
(factor resulting denominator) (radiation)
Answers to Practice Exercises
1. 3cx1x2 - 32 2. 3cx13x2 - 12
3. 13c - 82 13c + 82 4. 213c - 82 13c + 82
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CHAPTER 6 Factoring and Fractions
6.3
Factoring Trinomials
8IFO$PFGGJDJFOUPG4RVBSF5FSNJT t
*NQPSUBODFPGUIF.JEEMF5FSN t
'BDUPSJOH(FOFSBM5SJOPNJBMT t
Factoring by Grouping
In the previous section, we considered factoring based on the special products of
Eqs. (6.1) and (6.2). The trinomials formed from Eqs. (6.3) to (6.6) are important
expressions to be factored, and this section is devoted to them.
FACTORING TRINOMIALS FOR WHICH COEFFICIENT OF x2 IS 1
When factoring is based on Eq. (6.5), we start with the expression on the right and then
find the factors on the left. By writing Eq. (6.5) with sides reversed, we have
For reference, Eq. (6.5) is
1 x + a2 1 x + b2 = x2 + 1 a + b2 x + ab.
LEARNING TIP
When factoring trinomials of the
form x 2 + bx + c, we may be able to
find the two numbers that multiply
to c, and add to b, giving the form:
x 2 + bx + c = (x + number1) *
(x + number2)
sum
coefficient = 1
2
x + 1a + b2x + ab = 1x + a2 1x + b2
product
We find integers a and b by noting that
1. the coefficient of x2 is 1,
2. the final constant is the product of the constants a and b in the factors, and
3. the coefficient of x is the sum of a and b.
As in Section 6.2, we consider only factors in which all terms have integral
coefficients.
E X A M P L E 1 Factoring the trinomial x 2 + (a + b)x + ab
In factoring x2 + 3x + 2, we set it up as
x2 + 3x + 2 = 1x
sum
product
2 1x
2
integers
The constant 2 tells us that the product of the required integers is 2. Thus, the only possibilities are 2 and 1 (or 1 and 2). The + sign before the 2 indicates that the sign before
the 1 and 2 in the factors must be the same. The + sign before the 3, the sum of the
integers, tells us that both signs are positive. Therefore,
x2 + 3x + 2 = 1x + 22 1x + 12
In factoring x2 - 3x + 2, the analysis is the same until we note that the middle term
is negative. This tells us that both signs are negative in this case. Therefore,
x2 - 3x + 2 = 1x - 22 1x - 12
For a trinomial with first term x2 and constant +2 to be factorable, the middle term
must be +3x or -3x. No other middle terms are possible. This means, for example, the
expressions x2 + 4x + 2 and x2 - x + 2 cannot be factored.
■
E X A M P L E 2 Factoring trinomials with x 2 coefficient of 1
(a) In order to factor x2 + 7x - 8, we must find two integers whose product is -8 and
whose sum is +7. The possible factors of -8 are
-8 and +1
+8 and -1
-4 and +2
+4 and -2
Inspecting these, we see that only +8 and -1 have the sum of +7. Therefore,
LEARNING TIP
Always multiply the factors together
to check to see that you get the correct middle term.
x2 + 7x - 8 = 1x + 82 1x - 12
In choosing the correct values for the integers, it is usually fairly easy to find a
pair for which the product is the final term. However, choosing the pair of integers
that correctly fits the middle term is the step that often is not done properly. Special
attention must be given to choosing the integers so that the expansion of the resulting factors has the correct middle term of the original expression.
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6.3 Factoring Trinomials
191
(b) In the same way, we have
x2 - x - 12 = 1x - 42 1x + 32
since -4 and +3 is the only pair of integers whose product is -12 and whose sum
is -1.
COMMON ERROR
Ensure that you have the proper signs on the two numbers chosen to make the product
and the sum leading to the trinomial coefficients. To factor x2 - x - 6, we require two
numbers that multiply to -6. The pairs +2 and -3 and -2 and +3 both work. However,
only one of those pairs adds to the coefficient of the middle term (+2 and -3). Therefore,
x 2 - x - 6 = 1x + 2 2 1x - 3 2
x 2 - x - 6 ≠ 1x - 2 2 1x + 3 2
(c) Also,
x2 - 5xy + 6y2 = 1x - 3y2 1x - 2y2
since -3 and -2 is the only pair of integers whose product is +6 and whose sum is
-5. Here, we find second terms of each factor with a product of 6y2 and sum of
-5xy, which means that each second term must have a factor of y, as we have
shown above.
■
■ For reference, Eqs. (6.3) and (6.4) are
1 x + y 2 2 = x 2 + 2xy + y 2 and
1 x - y 2 2 = x2 - 2xy + y2.
In factoring a trinomial in which the first and third terms are perfect squares, we
may find that the expression fits the form of Eq. (6.3) or Eq. (6.4), as well as the form
of Eq. (6.5). The following example illustrates this case.
E X A M P L E 3 Factoring perfect square trinomials
(a) To factor x2 + 10x + 25, we must find two integers whose product is +25 and
whose sum is +10. Since 52 = 25, we note that this expression may fit the form of
Eq. (6.3). This can be the case only if the first and third terms are perfect squares.
Since the sum of +5 and +5 is +10, we have
or
Practice Exercises
Factor: 1. x2 - x - 2
2. x2 - 4x + 4
x2 + 10x + 25 = 1x + 52 1x + 52
x2 + 10x + 25 = 1x + 52 2
(b) To factor A4 - 20A2 + 100, we note that A4 = 1A2 2 2 and 100 = 102. This means
that this expression may fit the form of Eq. (6.4). Since the sum of -10 and -10 is
-20, we have
A4 - 20A2 + 100 = 1A2 - 102 1A2 - 102
= 1A2 - 102 2
(c) Just because the first and third terms are perfect squares, we must realize that the
expression may be factored but does not fit the form of either Eq. (6.3) or (6.4).
One example, using x2 and 100 for the first and third terms, is
x2 + 29x + 100 = 1x + 252 1x + 42
■
FACTORING GENERAL TRINOMIALS
Factoring a trinomial for which the coefficient of x2 is any positive integer involves a
slightly more challenging method. Factoring based on the special product of Eq. (6.6)
often requires some trial and error. However, the amount of trial and error can be kept
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CHAPTER 6 Factoring and Fractions
■ For reference, Eq. (6.6) is
1 ax + b2 1 cx + d2 =
acx2 + 1 ad + bc 2 x + bd.
to a minimum by carefully noting the coefficients of x2 and the constant. Rewriting
Eq. (6.6) with sides reversed, we have
product
coefficients
2
acx + 1ad + bc2x + bd = 1ax + b2 1cx + d2
integers
product
outer product
2
acx + 1ad + bc2x + bd = 1ax + b2 1cx + d2
inner product
This diagram shows us that
1. the coefficient of x2 is the product of the coefficients a and c in the factors,
2. the final constant is the product of the constants b and d in the factors, and
3. the coefficient of x is the sum of the inner and outer products.
In finding the factors, we must try possible combinations of a, b, c, and d that give
the proper inner and outer products for the middle term
of the given expression.
E X A M P L E 4 Factoring a general trinomial
To factor 2x2 + 11x + 5, we take the factors of 2 to be +2 and +1 (we use only positive
coefficients a and c when the coefficient of x2 is positive). We set up the factoring as
2x2 + 11x + 5 = 12x
2 1x
2
Since the product of the integers to be found is +5, only integers of the same sign need
to be considered. Also since the sum of the outer and inner products is +11, the integers are positive. The factors of +5 are +1 and +5, and -1 and -5, which means that
+1 and +5 is the only possible pair. Now, trying the factors
12x + 52 1x + 12
+ 5x
+ 2x
+ 2x + 5x = + 7x
we see that 7x is not the correct middle term. Therefore, we now try
12x + 12 1x + 52
+x
+ 10x
+ x + 10x = + 11x
and we have the correct sum of +11x. Therefore,
2x2 + 11x + 5 = 12x + 12 1x + 52
For a trinomial with a first term 2x2 and a constant +5 to be factorable, we can now
see that the middle term must be either +11x or +7x. This means that 2x2 + 7x + 5 =
12x + 52 1x + 12, but a trinomial such as 2x2 + 8x + 5 is not factorable.
■
E X A M P L E 5 Factoring a general trinomial with multiple leading
coefficient factors
In factoring 4x2 + 4x - 3, the coefficient 4 in 4x2 shows that the possible coefficients
of x in the factors are 4 and 1, or 2 and 2. The 3 shows that the only possible constants
in the factors are 1 and 3, and the minus sign with the 3 tells us that these integers have
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6.3 Factoring Trinomials
193
different signs. This gives us the following possible combinations of factors, along
with the resulting sum of the outer and inner products:
LEARNING TIP
Another hint given by the coefficients of the original expression is
that the plus sign on the x-term tells
us that the larger of the outer and
inner product factors must be
positive.
(4x + 3)(x - 1): -4x + 3x = -x
(4x + 1)(x - 3): -12x + x = -11x
14x - 32 1x + 12: 4x - 3x = +x
14x - 12 1x + 32: 12x - x = +11x
12x + 32 12x - 12: -2x + 6x = +4x
12x - 32 12x + 12: 2x - 6x = -4x
We see that the factors that have the correct middle term of +4x are 12x + 32 12x - 12.
This means that
4x2 + 4x - 3 = 12x + 32 12x - 12
+ 6x
- 2x
- 2x + 6x = + 4x
Expressing the result with the factors reversed is an equally correct answer.
■
E X A M P L E 6 Factoring a general trinomial with multiple factor
combinations on the x2-term and the last term
6s2 + 19st - 20t 2 = 16s - 5t2 1s + 4t2
+ 24st - 5st = + 19st
There are numerous possibilities for the combinations of 6 and −20. However, we must
remember to check carefully that the middle term of the expression is the proper
result of the factors we have chosen.
■
E X A M P L E 7 Check the middle term carefully
(a) In factoring 9x2 - 6x + 1, note that 9x2 is the square of 3x and 1 is the square of
1. Therefore, we recognize that this expression might fit the perfect square form of
Eq. (6.4). This leads us to factor it tentatively as
Practice Exercise
Factor: 3. 4x2 + x - 5
9x2 - 6x + 1 = 13x - 12 2
check middle term: 2 13x 2 1 - 1 2 = - 6x
However, before we can be certain that this is correct, we must check to see if the
middle term of the expansion of 13x - 12 2 is -6x, which is what it must be to fit
the form of Eq. (6.4). When we expand 13x - 12 2, we find that the middle term
is -6x and therefore that the factorization is correct.
(b) In the same way, we have
36x2 + 84xy + 49y2 = 16x + 7y2 2
check middle term: 2 16x 217y 2 = 84xy
since the middle term of the expansion of 16x + 7y2 2 is 84xy. We must be careful
to include the factors of y in the second terms of the factors.
■
COMMON ERROR
When the last term in a trinomial is not just a constant but also has a variable dependence, it is common to forget this variable when stating the factors:
2x 2 + 5xy - 3y 2 = 12x - y 2 1x + 3y)
2x 2 + 5xy - 3y 2 ≠ 12x - 1 2 1x + 3 2
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CHAPTER 6 Factoring and Fractions
As noted before, we must be careful to factor an expression completely. We first
look for common monomial factors and then check each resulting factor to see if it can
be factored when we complete each step.
E X A M P L E 8 General trinomial with common monomial factor
When factoring 2x2 + 6x - 8, first note the common monomial factor of 2. This leads to
2x2 + 6x - 8 = 21x2 + 3x - 42
Now, notice that x2 + 3x - 4 is also factorable. Therefore,
2x2 + 6x - 8 = 21x + 42 1x - 12
Having noted the common factor of 2 prevents our having to check factors of 2 and
8. If we had not noted the common factor, we might have arrived at
12x + 82 1x - 12
or
12x - 22 1x + 42
(possibly after a number of trials). Each is correct as far as it goes, but also each of
these is not complete. Since 2x + 8 = 21x + 42, or 2x - 2 = 21 x - 12 , we can
arrive at the proper result shown above. In this way, we would have
Practice Exercise
2
Factor: 4. 6x + 9x - 6
or
2x2 + 6x - 8 = 12x + 82 1x - 12
= 21x + 42 1x - 12
2x2 + 6x - 8 = 12x - 22 1x + 42
= 21x - 12 1x + 42
Although these factorizations are correct, it is more efficient to factor out the common
factor first. By doing so, the number of possible factoring combinations is greatly
reduced.
■
E X A M P L E 9 Factoring a trinomial—application
A study of the path of a certain rocket leads to the expression 16t 2 + 240t - 1600,
where t is the time of flight. Factor this expression.
An inspection shows that there is a common factor of 16. (This might be found by
noting successive factors of 2 or 4.) Factoring out 16 leads to
■ Liquid-fuel rockets were designed in the
United States in the 1920s but were developed
by German engineers. They were first used in
the 1940s during World War II.
16t 2 + 240t - 1600 = 161t 2 + 15t - 1002
= 161t + 202 1t - 52
numbers that multiply
to - 100 and sum to 15
are 20 and - 5
Here, factors of 100 need to be checked for sums equal to 15. This might take some time,
but it is much simpler than looking for factors of 16 and 1600 with sums of 240.
■
FACTORING BY GROUPING
Just as in Example 9 of Section 6.2, we can sometimes factor trinomials by grouping
terms in the expression together and finding common factors between them. To factor
the trinomial ax2 + bx + c by grouping, we use the following procedure.
Method for Factoring a Trinomial by Grouping
1. Find two numbers whose product is ac and whose sum is b.
2. Write the trinomial with two x-terms having these numbers as coefficients.
3. Complete the factorization by grouping.
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6.3 Factoring Trinomials
195
E X A M P L E 1 0 Factoring by grouping
Factor the trinomial 6x2 + 7x - 20 by grouping.
We begin by finding two numbers with a product of -120 and a sum of 7. Trying
products of numbers with different signs and a sum of 7, we find the numbers that
work are -8 and 15. We then complete steps 2 and 3 of the above procedure as
follows:
6x2 + 7x - 20 =
=
=
=
6x2 - 8x + 15x - 20
16x2 - 8x2 + 115x - 202
2x13x - 42 + 513x - 42
13x - 42 12x + 52
7x = - 8x + 15x
group first two terms and last two terms
find common factor of each group and
note common factor of 3x - 4
Multiplication verifies that these are the correct factors.
■
E X A M P L E 1 1 Factoring by grouping
Factor x2 - 4xy + 4y2 - 9.
The first three terms represent 1x - 2y2 2. Thus, grouping these terms, we have
x2 - 4xy + 4y2 - 9 =
=
=
=
1x2 - 4xy + 4y2 2 - 9
group terms
2
1x - 2y2 - 9
factor grouping (note difference of squares)
3 1 x - 2y2 + 34 3 1 x - 2y2 - 34 factor difference of squares
1x - 2y + 32 1x - 2y - 32
Not all groupings work. Noting the factorable combination 4y2 - 9, and then grouping the first two terms and the last two terms, would not have led to the factorization. ■
E XE RC IS ES 6 .3
In Exercises 1–6, make the given changes in the indicated examples of
this section and then factor.
1. In Example 1, change the 3 to 4 and the 2 to 3.
2
41. 4p - 25pq + 6q
8. x2 - 5x - 6
9. s2 - s - 42
10. a2 + 14a - 32
14. D2 + 8D + 16
2
16. b - 12bc + 36c
2
20. 25x + 45x - 10
2
27. 3t - 7tu + 4u
28. 3x + xy - 14y
29. 4x2 - 3x - 7
30. 2z2 + 13z - 5
2
31. 9x + 7xy - 2y
2
2
2
57. 16t 2 - 80t + 64
(projectile motion)
58. 9x2 - 33Lx + 30L2
2
(civil engineering)
(electricity)
62. bT 2 - 40bT + 400b
63. V 2 - 2nBV + n2B2
2
32. 4r + 11rs - 3s
54. r 2 - s2 + 2st - t 2
In Exercises 57–68, factor the given expressions completely. Each is
from the technical area indicated.
(fuel efficiency)
61. 200n - 2100n - 3600
26. 3n - 20n + 20
2
2
56. 12B2n + 19BnH - 10H 2
2
2
2
52. x2 - 6xy + 9y2 - 4z2
2
60. 3h2 + 18h - 1560
2
24. 5R - 3R - 2
25. 2t + 7t - 15
50. 6x4 - 13x3 + 5x2
49. ax + 4a x - 12a x
59. 4s + 16s + 12
22. 7y - 12y + 5
2
3
55. 4x2n + 13xn - 12
18. 2n2 - 13n - 7
23. 3f - 16f + 5
2
2
4
48. 12B2 + 22BH - 4H 2
2 2
53. 25a - 25x - 10xy - y
13. x2 + 2x + 1
2
3
2
12. r 3 - 11r 2 + 18r
4
46. 6y2 - 33y - 18
51. a2 + 2ab + b2 - 4
11. t 2 + 5t - 24
21. 2s + 13s + 11
44. 8r 2 - 14rs - 9s2
47. 4x5 + 14x3 - 8x
7. x2 + 5x + 4
2
42. 12x2 + 4xy - 5y2
45. 2x - 14x + 12
In Exercises 7–56, factor the given expressions completely.
19. 12y - 32y - 12
40. 12n4 + 8n2 - 15
2
2
6. In Example 10, change the + before 7x to -.
2
38. 6t 4 + t 2 - 12
43. 12x2 + 47xy - 4y2
5. In Example 8, change the 8 to 36.
17. 3x2 - 5x - 2
36. 3a2c2 - 6ac + 3
39. 8b6 + 31b3 - 4
4. In Example 6, change the coefficient 19 to 7.
15. L - 4LK + 4K
35. 8x2 - 24x + 18
37. 9t - 15t + 4
3. In Example 4, change the + before 11x to -.
2
34. 16q2 + 24q + 9
2
2. In Example 2(a), change the + before 7x to -.
2
33. 4m2 + 20m + 25
(biology)
(thermodynamics)
(chemistry)
64. a4 + 8a2p2f 2 + 16p4f 4
(periodic motion: energy)
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196
CHAPTER 6 Factoring and Fractions
65. wx4 - 5wLx3 + 6wL2x2
66. 1 - 2r 2 + r 4
72. Using the method of Exercise 71, factor x4 + x2 + 1. (First add
and subtract x2.)
(beam design)
(lasers)
67. 3Adu2 - 4Aduv + Adv 2
(water power)
68. k 2A2 + 2klA + l2 - a2
(robotics)
73. If sum of squares 36x2 + 9 can be factored, factor it.
74. Explain why most students would find 24x2 - 23x - 12 more
difficult to factor than 23x2 - 18x - 5.
In Exercises 69–75, solve the given problems.
75. Factor the statements in Exercises 40–43 using grouping.
69. Find the integral value of k that makes 4x2 + 4x - k a perfect
square trinomial, and express the result in factored form.
70. Find the integral value of k that makes 49x2 - 70x + k a perfect
square trinomial, and express the result in factored form.
4
71. Often, the sum of squares cannot be factored, but x + 4 can be
factored. Add and subtract 4x2 and then use factoring by grouping.
6.4
Answers to Practice Exercises
1. 1x - 22 1x + 12
3. 14x + 52 1x - 12
2. 1x - 22 2
4. 312x - 12 1x + 22
The Sum and Difference of Cubes
'BDUPSJOHUIF4VNPG$VCFT t 'BDUPSJOH
UIF%JGGFSFODFPG$VCFT t 4VNNBSZPG
Methods of Factoring
We have seen that the difference of squares can be factored, but that the sum of squares
often cannot be factored. We now turn our attention to the sum and difference of cubes,
both of which can be factored. Writing Eqs. (6.9) and (6.10) with sides reversed,
■ Note carefully the positions of the
+ and – signs.
x3 + y3 = 1x + y2 1x2 - xy + y2 2
x3 - y3 = 1x - y2 1x2 + xy + y2 2
(6.9)
(6.10)
In these equations, the second factors are prime, assuming x and y are themselves prime
factors.
E X A M P L E 1 Factoring the sum and difference of cubes
(b) x3 - 1 = x3 - 13
3
Table of Cubes
13 =
1
23 =
8
3
3 =
27
43 =
64
(a) x + 8 = x3 + 23
= 1x + 22 3 1x2 2 - 2x + 22 4
= 1x + 22 1x2 - 2x + 42
(c) 8 - 27x3 = 23 - 13x2 3
= 12 - 3x2 3 22 + 213x2 + 13x2 2 4
= 12 - 3x2 14 + 6x + 9x2 2
53 = 125
63 = 216
= 1x - 1231x2 2 + 1121x2 + 124
= 1x - 121x2 + x + 12
8 = 23 and 27x3 = 13x 2 3
■
E X A M P L E 2 Factoring a difference of cubes with an initial common factor
In factoring ax5 - ax2, we first note that each term has a common factor of ax2. This is
factored out to get ax2 1x3 - 12. However, the expression is not completely factored
since 1 = 13, which means that x3 - 1 is the difference of cubes. We complete the
factoring by the use of Eq. (6.10). Therefore,
Practice Exercises
3
Factor: 1. x + 216
3
2. x - 216
ax5 - ax2 = ax2 1x3 - 12
= ax2 1x - 12 1x2 + x + 12
■
E X A M P L E 3 Factoring a difference of cubes—application
The volume of material used to make a steel bearing with a hollow core is given by
4
4
3
3
3 pR - 3 pr . Factor this expression.
4 3
4
4
pR - pr 3 = p1R3 - r 3 2
3
3
3
4
= p1R - r 2 1R2 + Rr + r 2 2
3
common factor of
4
p
3
using Eq. (6.10) ■