APPENDIX
Partial
Fractions
Outline
Basic Theorems
Partial Fraction Decomposition
B
You have now had considerable experience combining two or more rational
expressions into a single rational expression. For example, problems such as
2
3
2(x ⫺ 4) ⫹ 3(x ⫹ 5)
5x ⫹ 7
⫹
⫽
⫽
x⫹5 x⫺4
(x ⫹ 5)(x ⫺ 4)
(x ⫹ 5)(x ⫺ 4)
should seem routine. Frequently in more advanced courses, particularly in calculus, it is advantageous to be able to reverse this process—that is, to be able to
express a rational expression as the sum of two or more simpler rational expressions called partial fractions. As is often the case with reverse processes, the
process of decomposing a rational expression into partial fractions is more difficult than combining rational expressions. Basic to the process is the factoring of
polynomials, so the topics discussed in Section 3-2 can be put to effective use.
We confine our attention to rational expressions of the form P(x)/D(x), where
P(x) and D(x) are polynomials with real coefficients. In addition, we assume that
the degree of P(x) is less than the degree of D(x). If the degree of P(x) is greater
than or equal to that of D(x), we have only to divide P(x) by D(x) to obtain
R(x)
P(x)
⫽ Q(x) ⫹
D(x)
D(x)
where the degree of R(x) is less than that of D(x). For example,
x 4 ⫺ 3x3 ⫹ 2x2 ⫺ 5x ⫹ 1
⫺6x ⫹ 2
⫽ x2 ⫺ x ⫺ 1 ⫹ 2
x2 ⫺ 2x ⫹ 1
x ⫺ 2x ⫹ 1
If the degree of P(x) is less than that of D(x), then P(x)/D(x) is called a proper
fraction.
Basic Theorems
Our task now is to establish a systematic way to decompose a proper fraction into
the sum of two or more partial fractions. The following three theorems take care
of the problem completely. Theorems 1 and 3 are stated without proof.
THEOREM
1
EQUAL POLYNOMIALS
Two polynomials are equal to each other if and only if the coefficients of
terms of like degree are equal.
For example, if
Equate the constant terms.
(A ⫹ 2B)x ⫹ B ⫽ 5x ⫺ 3
afddddbddddfc
Equate the coefficients of x.
A-84
Appendix B
PARTIAL FRACTIONS
A-85
then
B ⫽ ⫺3
A ⫹ 2B ⫽ 5
Substitute B ⫽ ⫺3 into the second equation to
solve for A.
A ⫹ 2(ⴚ3) ⫽ 5
A ⫽ 11
Explore/Discuss
1
If
x ⫹ 5 ⫽ A(x ⫹ 1) ⫹ B(x ⫺ 3)
(1)
is a polynomial identity (that is, both sides represent the same polynomial), then equating coefficients produces the system
1⫽A⫹ B
Equating coefficients of x
5 ⫽ A ⫺ 3B
Equating constant terms
(A) Solve this system graphically.
(B) For an alternate method of solution, substitute x ⫽ 3 in equation (1)
to find A and then substitute x ⫽ ⫺1 in equation (1) to find B.
Explain why this method is valid.
THEOREM
2
LINEAR AND QUADRATIC FACTOR THEOREM
For a polynomial with real coefficients, there always exists a complete
factoring involving only linear and/or quadratic factors with real coefficients where the linear and quadratic factors are prime relative to the real
numbers.
That Theorem 2 is true can be seen as follows: From earlier theorems in Chapter 3, we know that an nth-degree polynomial P(x) has n zeros and n linear factors. The real zeros of P(x) correspond to linear factors of the form (x ⫺ r), where
r is a real number. Since P(x) has real coefficients, the imaginary zeros occur in
conjugate pairs. Thus, the imaginary zeros correspond to pairs of factors of the
form [x ⫺ (a ⫹ bi)] and [x ⫺ (a ⫺ bi)], where a and b are real numbers. Multiplying these two imaginary factors, we have
[x ⫺ (a ⫹ bi)][x ⫺ (a ⫺ bi)] ⫽ x2 ⫺ 2ax ⫹ a2 ⫹ b2
This quadratic polynomial with real coefficients is a factor of P(x). Thus, P(x)
can be factored into a product of linear factors and quadratic factors, all with real
coefficients.
A-86
Appendix B
PARTIAL FRACTIONS
Partial Fraction Decomposition
We are now ready to state Theorem 3, which forms the basis for partial fraction
decomposition.
THEOREM
3
PARTIAL FRACTION DECOMPOSITION
Any proper fraction P(x)/D(x) reduced to lowest terms can be decomposed into the sum of partial fractions as follows:
1. If D(x) has a nonrepeating linear factor of the form ax ⫹ b, then the
partial fraction decomposition of P(x)/D(x) contains a term of the
form
A
ax ⫹ b
A a constant
2. If D(x) has a k-repeating linear factor of the form (ax ⫹ b)k, then the
partial fraction decomposition of P(x)/D(x) contains k terms of the
form
A2
Ak
A1
⫹
⫹...⫹
2
ax ⫹ b (ax ⫹ b)
(ax ⫹ b)k
A1, A2, . . . , Ak constants
3. If D(x) has a nonrepeating quadratic factor of the form ax2 ⫹ bx ⫹
c, which is prime relative to the real numbers, then the partial fraction decomposition of P(x)/D(x) contains a term of the form
Ax ⫹ B
ax ⫹ bx ⫹ c
2
A, B constants
4. If D(x) has a k-repeating quadratic factor of the form (ax2 ⫹ bx
⫹ c)k, where ax2 ⫹ bx ⫹ c is prime relative to the real numbers,
then the partial fraction decomposition of P(x)/D(x) contains k terms
of the form
A1x ⫹ B1
A2x ⫹ B2
Ak x ⫹ Bk
⫹
⫹...⫹
ax2 ⫹ bx ⫹ c (ax2 ⫹ bx ⫹ c)2
(ax2 ⫹ bx ⫹ c)k
A1, . . . , Ak, B1, . . . , Bk constants
Let’s see how the theorem is used to obtain partial fraction decompositions in
several examples.
EXAMPLE
Nonrepeating Linear Factors
1
Decompose into partial fractions:
5x ⫹ 7
.
x ⫹ 2x ⫺ 3
2
Appendix B
Solution
PARTIAL FRACTIONS
A-87
We first try to factor the denominator. If it can’t be factored in the real numbers,
then we can’t go any further. In this example, the denominator factors, so we
apply part 1 from Theorem 3:
5x ⫹ 7
A
B
⫽
⫹
(x ⫺ 1)(x ⫹ 3) x ⫺ 1 x ⫹ 3
(2)
To find the constants A and B, we combine the fractions on the right side of equation (2) to obtain
5x ⫹ 7
A(x ⫹ 3) ⫹ B(x ⫺ 1)
⫽
(x ⫺ 1)(x ⫹ 3)
(x ⫺ 1)(x ⫹ 3)
Since these fractions have the same denominator, their numerators must be equal.
Thus
5x ⫹ 7 ⫽ A(x ⫹ 3) ⫹ B(x ⫺ 1)
(3)
We could multiply the right side and find A and B by using Theorem 1, but in
this case it is easier to take advantage of the fact that equation (3) is an identity—that is, it must hold for all values of x. In particular, we note that if we let
x ⫽ 1, then the second term of the right side drops out and we can solve for A:
5 ⴢ 1 ⫹ 7 ⫽ A(1 ⫹ 3) ⫹ B(1 ⫺ 1)
12 ⫽ 4A
A⫽3
Similarly, if we let x ⫽ ⫺3, the first term drops out and we find
⫺8 ⫽ ⫺4B
B⫽2
Hence,
5x ⫹ 7
3
2
⫽
⫹
x ⫹ 2x ⫺ 3 x ⫺ 1 x ⫹ 3
(4)
2
as can easily be checked by adding the two fractions on the right.
MATCHED PROBLEM
1
Decompose into partial fractions:
7x ⫹ 6
.
x2 ⫹ x ⫺ 6
A-88
Appendix B
PARTIAL FRACTIONS
Explore/Discuss
2
A graphing utility can also be used to check a partial fraction decomposition. To check Example 1, we graph the left and right sides of equation
(4) in a graphing utility (Fig. 1). Discuss how the TRACE feature on the
graphing utility can be used to check that the graphing utility is displaying two identical graphs.
FIGURE 1
10
⫺10
10
⫺10
EXAMPLE
2
Solution
Repeating Linear Factors
Decompose into partial fractions:
6x2 ⫺ 14x ⫺ 27
.
(x ⫹ 2)(x ⫺ 3)2
Using parts 1 and 2 from Theorem 3, we write
6x2 ⫺ 14x ⫺ 27
A
B
C
⫹
⫹
⫽
2
(x ⫹ 2)(x ⫺ 3)
x ⫹ 2 x ⫺ 3 (x ⫺ 3)2
⫽
A(x ⫺ 3)2 ⫹ B(x ⫹ 2)(x ⫺ 3) ⫹ C(x ⫹ 2)
(x ⫹ 2)(x ⫺ 3)2
Thus, for all x,
6x2 ⫺ 14x ⫺ 27 ⫽ A(x ⫺ 3)2 ⫹ B(x ⫹ 2)(x ⫺ 3) ⫹ C(x ⫹ 2)
If x ⫽ 3, then
If x ⫽ ⫺2, then
⫺15 ⫽ 5C
25 ⫽ 25A
C ⫽ ⫺3
A⫽1
There are no other values of x that will cause terms on the right to drop out. Since
any value of x can be substituted to produce an equation relating A, B, and C, we
let x ⫽ 0 and obtain
⫺27 ⫽ 9A ⫺ 6B ⫹ 2C
⫺27 ⫽ 9 ⫺ 6B ⫺ 6
B⫽5
Substitute A ⫽ 1 and C ⫽ ⫺3.
Appendix B
PARTIAL FRACTIONS
A-89
Thus,
6x2 ⫺ 14x ⫺ 27
1
5
3
⫽
⫹
⫺
2
(x ⫹ 2)(x ⫺ 3)
x ⫹ 2 x ⫺ 3 (x ⫺ 3)2
MATCHED PROBLEM
Decompose into partial fractions:
2
EXAMPLE
3
Solution
x2 ⫹ 11x ⫹ 15
.
(x ⫺ 1)(x ⫹ 2)2
Nonrepeating Linear and Quadratic Factors
Decompose into partial fractions:
5x2 ⫺ 8x ⫹ 5
.
(x ⫺ 2)(x2 ⫺ x ⫹ 1)
First, we see that the quadratic in the denominator can’t be factored further in the
real numbers. Then, we use parts 1 and 3 from Theorem 3 to write
5x2 ⫺ 8x ⫹ 5
A
Bx ⫹ C
⫽
⫹
(x ⫺ 2)(x2 ⫺ x ⫹ 1) x ⫺ 2 x2 ⫺ x ⫹ 1
⫽
A(x2 ⫺ x ⫹ 1) ⫹ (Bx ⫹ C)(x ⫺ 2)
(x ⫺ 2)(x2 ⫺ x ⫹ 1)
Thus, for all x,
5x2 ⫺ 8x ⫹ 5 ⫽ A(x2 ⫺ x ⫹ 1) ⫹ (Bx ⫹ C)(x ⫺ 2)
If x ⫽ 2, then
9 ⫽ 3A
A⫽3
If x ⫽ 0, then, using A ⫽ 3, we have
5 ⫽ 3 ⫺ 2C
C ⫽ ⫺1
If x ⫽ 1, then, using A ⫽ 3 and C ⫽ ⫺1, we have
2 ⫽ 3 ⫹ (B ⫺ 1)(⫺1)
B⫽2
Hence,
5x2 ⫺ 8x ⫹ 5
3
2x ⫺ 1
⫽
⫹ 2
2
(x ⫺ 2)(x ⫺ x ⫹ 1) x ⫺ 2 x ⫺ x ⫹ 1
A-90
Appendix B
PARTIAL FRACTIONS
MATCHED PROBLEM
Decompose into partial fractions:
3
7x2 ⫺ 11x ⫹ 6
.
(x ⫺ 1)(2x2 ⫺ 3x ⫹ 2)
EXAMPLE
Repeating Quadratic Factors
4
Decompose into partial fractions:
x3 ⫺ 4x2 ⫹ 9x ⫺ 5
.
(x2 ⫺ 2x ⫹ 3)2
Since x2 ⫺ 2x ⫹ 3 can’t be factored further in the real numbers, we proceed to
use part 4 from Theorem 3 to write
Solution
x3 ⫺ 4x2 ⫹ 9x ⫺ 5
Ax ⫹ B
Cx ⫹ D
⫽ 2
⫹
(x2 ⫺ 2x ⫹ 3)2
x ⫺ 2x ⫹ 3 (x2 ⫺ 2x ⫹ 3)2
⫽
(Ax ⫹ B)(x2 ⫺ 2x ⫹ 3) ⫹ Cx ⫹ D
(x2 ⫺ 2x ⫹ 3)2
Thus, for all x,
x3 ⫺ 4x2 ⫹ 9x ⫺ 5 ⫽ (Ax ⫹ B)(x2 ⫺ 2x ⫹ 3) ⫹ Cx ⫹ D
Since the substitution of carefully chosen values of x doesn’t lead to the immediate determination of A, B, C, or D, we multiply and rearrange the right side to obtain
x3 ⫺ 4x2 ⫹ 9x ⫺ 5 ⫽ Ax3 ⫹ (B ⫺ 2A)x2 ⫹ (3A ⫺ 2B ⫹ C)x ⫹ (3B ⫹ D)
Now we use Theorem 1 to equate coefficients of terms of like degree:
⫺5
abc
abc
abc
3B ⫹ D ⫽ ⫺5
⫹9x
abc
afddddbddddfc
3A ⫺ 2B ⫹ C ⫽ 9
⫺4x 2
afddddgddddbddddddddgfc
B ⫺ 2A ⫽ ⫺4
1x 3
abc
afddddbddddfc
A⫽1
Ax 3 ⫹ (B ⫺ 2A)x 2 ⫹ (3A ⫺ 2B ⫹ C )x ⫹ (3B ⫹ D)
From these equations we easily find that A ⫽ 1, B ⫽ ⫺2, C ⫽ 2, and D ⫽ 1.
Now we can write
x3 ⫺ 4x2 ⫹ 9x ⫺ 5
x⫺2
2x ⫹ 1
⫽ 2
⫹ 2
2
2
(x ⫺ 2x ⫹ 3)
x ⫺ 2x ⫹ 3 (x ⫺ 2x ⫹ 3)2
MATCHED PROBLEM
Decompose into partial fractions:
4
3x3 ⫺ 6x2 ⫹ 7x ⫺ 2
.
(x2 ⫺ 2x ⫹ 2)2
Answers to Matched Problems
1.
4
3
⫹
x⫺2 x⫹3
2.
3
2
1
⫺
⫹
x ⫺ 1 x ⫹ 2 (x ⫹ 2)2
3.
2
3x ⫺ 2
⫹
x ⫺ 1 2x2 ⫺ 3x ⫹ 2
4.
3x
x⫺2
⫹
x2 ⫺ 2x ⫹ 2 (x2 ⫺ 2x ⫹ 2)2
Appendix B
EXERCISE B
A
In Problems 1–10, find constants A, B, C, and D so that the
right side is equal to the left.
PARTIAL FRACTIONS
13.
3x ⫺ 13
6x2 ⫺ x ⫺ 12
14.
11x ⫺ 11
6x2 ⫹ 7x ⫺ 3
15.
x2 ⫺ 12x ⫹ 18
x3 ⫺ 6x2 ⫹ 9x
16.
5x2 ⫺ 36x ⫹ 48
x(x ⫺ 4)2
17.
5x2 ⫹ 3x ⫹ 6
x3 ⫹ 2x2 ⫹ 3x
18.
6x2 ⫺ 15x ⫹ 16
x3 ⫺ 3x2 ⫹ 4x
A-91
1.
7x ⫺ 14
A
B
⫽
⫹
(x ⫺ 4)(x ⫹ 3) x ⫺ 4 x ⫹ 3
19.
20.
⫺5x2 ⫹ 7x ⫺ 18
x 4 ⫹ 6x2 ⫹ 9
2.
9x ⫹ 21
A
B
⫽
⫹
(x ⫹ 5)(x ⫺ 3) x ⫹ 5 x ⫺ 3
2x3 ⫹ 7x ⫹ 5
x 4 ⫹ 4x2 ⫹ 4
21.
22.
x3 ⫹ x2 ⫺ 13x ⫹ 11
x2 ⫹ 2x ⫺ 15
3.
17x ⫺ 1
A
B
⫽
⫹
(2x ⫺ 3)(3x ⫺ 1) 2x ⫺ 3 3x ⫺ 1
x3 ⫺ 7x2 ⫹ 17x ⫺ 17
x2 ⫺ 5x ⫹ 6
4.
x ⫺ 11
A
B
⫽
⫹
(3x ⫹ 2)(2x ⫺ 1) 3x ⫹ 2 2x ⫺ 1
5.
B
C
3x2 ⫹ 7x ⫹ 1 A
⫽ ⫹
⫹
x(x ⫹ 1)2
x
x ⫹ 1 (x ⫹ 1)2
x2 ⫺ 6x ⫹ 11
A
B
C
⫽
⫹
⫹
6.
(x ⫹ 1)(x ⫺ 2)2 x ⫹ 1 x ⫺ 2 (x ⫺ 2)2
C
In Problems 23–30, decompose into partial fractions.
23.
4x2 ⫹ 5x ⫺ 9
x3 ⫺ 6x ⫺ 9
4x2 ⫺ 8x ⫹ 1
x3 ⫺ x ⫹ 6
7.
3x ⫹ x
A
Bx ⫹ C
⫽
⫹ 2
(x ⫺ 2)(x2 ⫹ 3) x ⫺ 2
x ⫹3
24.
8.
5x ⫺ 9x ⫹ 19
A
Bx ⫹ C
⫽
⫹ 2
(x ⫺ 4)(x2 ⫹ 5) x ⫺ 4
x ⫹5
25.
x2 ⫹ 16x ⫹ 18
x3 ⫹ 2x2 ⫺ 15x ⫺ 36
26.
5x2 ⫺ 18x ⫹ 1
x ⫺ x2 ⫺ 8x ⫹ 12
27.
⫺x2 ⫹ x ⫺ 7
x ⫺ 5x3 ⫹ 9x2 ⫺ 8x ⫹ 4
28.
⫺2x3 ⫹ 12x2 ⫺ 20x ⫺ 10
x ⫺ 7x3 ⫹ 17x2 ⫺ 21x ⫹ 18
29.
4x5 ⫹ 12x4 ⫺ x3 ⫹ 7x2 ⫺ 4x ⫹ 2
4x 4 ⫹ 4x3 ⫺ 5x2 ⫹ 5x ⫺ 2
30.
6x5 ⫺ 13x 4 ⫹ x3 ⫺ 8x2 ⫹ 2x
6x 4 ⫺ 7x3 ⫹ x2 ⫹ x ⫺ 1
2
2
2x2 ⫹ 4x ⫺ 1
Ax ⫹ B
Cx ⫹ D
⫽ 2
⫹ 2
9. 2
2
(x ⫹ x ⫹ 1)
x ⫹ x ⫹ 1 (x ⫹ x ⫹ 1)2
Ax ⫹ B
Cx ⫹ D
3x ⫺ 3x ⫹ 10x ⫺ 4
⫽ 2
⫹
(x2 ⫺ x ⫹ 3)2
x ⫺ x ⫹ 3 (x2 ⫺ x ⫹ 3)2
3
10.
2
B
In Problems 11–22, decompose into partial fractions.
11.
⫺x ⫹ 22
x2 ⫺ 2x ⫺ 8
12.
⫺x ⫺ 21
x2 ⫹ 2x ⫺ 15
3
4
4