SECTION 10.5
10.5
Partial Fraction Decomposition
785
Partial Fraction Decomposition
PREPARING FOR THIS SECTION
Before getting started, review the following:
• Identity (Appendix, Section A.5 p. 984)
• Factoring Polynomials (Appendix, Section A.3, pp. 969–971)
• Proper and Improper Rational Functions (Section 3.3,
pp. 191–192)
• Fundamental Theorem of Algebra (Section 3.7, pp. 233–237)
Now work the ‘Are You Prepared?’ problems on page 792.
OBJECTIVES
1
2
3
4
P
, Where Q Has Only Nonrepeated Linear Factors
Q
P
Decompose , Where Q Has Repeated Linear Factors
Q
P
Decompose , Where Q Has a Nonrepeated Irreducible Quadratic Factor
Q
P
Decompose , Where Q Has Repeated Irreducible Quadratic Factors
Q
Decompose
Consider the problem of adding two rational expressions:
3
x + 4
and
2
x - 3
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The result is
31x - 32 + 21x + 42
3
2
5x - 1
+
=
= 2
x + 4
x - 3
1x + 421x - 32
x + x - 12
5x - 1
x + x - 12
3
and writing it as the sum (or difference) of the two simpler fractions
and
x + 4
2
, is referred to as partial fraction decomposition, and the two simpler fractions
x - 3
are called partial fractions. Decomposing a rational expression into a sum of partial
fractions is important in solving certain types of calculus problems. This section
presents a systematic way to decompose rational expressions.
We begin by recalling that a rational expression is the ratio of two polynomials, say,
P and Q Z 0. We assume that P and Q have no common factors. Recall also that a
P
rational expression
is called proper if the degree of the polynomial in the
Q
numerator is less than the degree of the polynomial in the denominator. Otherwise,
the rational expression is termed improper.
Because any improper rational expression can be reduced by long division to a
mixed form consisting of the sum of a polynomial and a proper rational expression,
we shall restrict the discussion that follows to proper rational expressions.
P
The partial fraction decomposition of the rational expression
depends on
Q
the factors of the denominator Q. Recall (from Section 3.7) that any polynomial whose
coefficients are real numbers can be factored (over the real numbers) into products
of linear and/or irreducible quadratic factors. This means that the denominator Q of
P
the rational expression
will contain only factors of one or both of the following
Q
types:
The reverse procedure, of starting with the rational expression
2
1. Linear factors of the form x - a, where a is a real number.
2. Irreducible quadratic factors of the form ax2 + bx + c, where a, b, and c are
real numbers, a Z 0, and b2 - 4ac 6 0 (which guarantees that ax2 + bx + c
cannot be written as the product of two linear factors with real coefficients).
As it turns out, there are four cases to be examined. We begin with the case for
which Q has only nonrepeated linear factors.
P
1 Decompose , Where Q Has Only Nonrepeated Linear Factors
✓
Q
Case 1: Q has only nonrepeated linear factors.
Under the assumption that Q has only nonrepeated linear factors, the
polynomial Q has the form
Q1x2 = 1x - a121x - a22 # Á # 1x - an2
where none of the numbers a1 , a2 , Á , an is equal. In this case, the partial
P
fraction decomposition of is of the form
Q
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SECTION 10.5
Partial Fraction Decomposition
P1x2
An
A1
A2
=
+
+ Á +
x - a1
x - a2
x - an
Q1x2
787
(1)
where the numbers A 1 , A 2 , Á , A n are to be determined.
We show how to find these numbers in the example that follows.
EXAMPLE 1
Nonrepeated Linear Factors
Write the partial fraction decomposition of
Solution
x
x - 5x + 6
2
First, we factor the denominator,
x2 - 5x + 6 = 1x - 221x - 32
and conclude that the denominator contains only nonrepeated linear factors. Then
we decompose the rational expression according to equation (1):
A
B
x
=
+
x - 2
x - 3
x - 5x + 6
2
(2)
where A and B are to be determined. To find A and B, we clear the fractions by
multiplying each side by 1x - 221x - 32 = x2 - 5x + 6. The result is
x = A1x - 32 + B1x - 22
or
(3)
x = 1A + B2x + 1-3A - 2B2
This equation is an identity in x. We equate the coefficients of like powers of x to get
b
1 =
A + B
0 = -3A - 2B
Equate coefficients of x: 1x = (A + B)x.
Equate coefficients of x0, the constants: 0x0 = (-3A - 2B)x0.
This system of two equations containing two variables, A and B, can be solved using
whatever method you wish. Solving it, we get
A = -2
B = 3
From equation (2), the partial fraction decomposition is
x
-2
3
=
+
x - 2
x - 3
x2 - 5x + 6
✔ CHECK: The decomposition can be checked by adding the rational expressions.
-21x - 32 + 31x - 22
-2
3
x
+
=
=
x - 2
x - 3
1x - 221x - 32
1x - 221x - 32
=
x
x2 - 5x + 6
The numbers to be found in the partial fraction decomposition can sometimes
be found more readily by using suitable choices for x (which may include complex
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numbers) in the identity obtained after fractions have been cleared. In Example 1,
the identity after clearing fractions is equation (3):
x = A1x - 32 + B1x - 22
If we let x = 2 in this expression, the term containing B drops out, leaving
2 = A1-12, or A = -2. Similarly, if we let x = 3, the term containing A drops out,
leaving 3 = B. As before, A = -2 and B = 3.
NOW WORK PROBLEM
13.
P
2 Decompose Where Q Has Repeated Linear Factors
✓
Q
Case 2:
Q has repeated linear factors.
If the polynomial Q has a repeated linear factor, say 1x - a2n, n Ú 2 an integer,
P
then, in the partial fraction decomposition of , we allow for the terms
Q
An
A1
A2
+
+ Á +
2
x - a
1x - a2n
1x - a2
where the numbers A 1 , A 2 , Á , A n are to be determined.
EXAMPLE 2
Repeated Linear Factors
Write the partial fraction decomposition of
Solution
x + 2
.
x - 2x2 + x
3
First, we factor the denominator,
x3 - 2x2 + x = x1x2 - 2x + 12 = x1x - 122
and find that the denominator has the nonrepeated linear factor x and the repeated
A
linear factor 1x - 122. By Case 1, we must allow for the term in the decomposition;
x
C
B
and, by Case 2, we must allow for the terms
in the decomposition.
+
x - 1
1x - 122
We write
x + 2
A
B
C
=
+
+
2
x
x
1
x - 2x + x
1x - 122
3
(4)
Again, we clear fractions by multiplying each side by x3 - 2x2 + x = x1x - 122.
The result is the identity
x + 2 = A1x - 122 + Bx1x - 12 + Cx
(5)
If we let x = 0 in this expression, the terms containing B and C drop out, leaving
2 = A1-122, or A = 2. Similarly, if we let x = 1, the terms containing A and B drop
out, leaving 3 = C. Then, equation (5) becomes
x + 2 = 21x - 122 + Bx1x - 12 + 3x
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SECTION 10.5
Partial Fraction Decomposition
789
Now let x = 2 (any choice other than 0 or 1 will work as well). The result is
4
4
2B
B
=
=
=
=
21122 + B122112 + 3122
2 + 2B + 6
-4
-2
We have A = 2, B = -2, and C = 3.
From equation (4), the partial fraction decomposition is
x + 2
-2
3
2
+
+
=
2
x
x
1
x - 2x + x
1x - 122
3
EXAMPLE 3
Repeated Linear Factors
Write the partial fraction decomposition of
Solution
x3 - 8
.
x 1x - 123
2
The denominator contains the repeated linear factor x2 and the repeated linear factor 1x - 123. The partial fraction decomposition takes the form
x3 - 8
A
B
C
D
E
=
+ 2 +
+
+
x
x - 1
x 1x - 123
x
1x - 122
1x - 123
2
(6)
As before, we clear fractions and obtain the identity
x3 - 8 = Ax1x - 123 + B1x - 123 + Cx21x - 122 + Dx21x - 12 + Ex2
(7)
Let x = 0. (Do you see why this choice was made?) Then
-8 = B1-12
B = 8
Now let x = 1 in equation (7). Then
-7 = E
Use B = 8 and E = -7 in equation (7) and collect like terms.
x3 - 8 = Ax1x - 123 + 81x - 123
+ Cx21x - 122 + Dx21x - 12 - 7x2
x - 8 - 81x - 3x + 3x - 12 + 7x = Ax1x - 123 + Cx21x - 122 + Dx21x - 12
3
3
2
2
-7x3 + 31x2 - 24x = x1x - 123A1x - 122 + Cx1x - 12 + Dx4
x1x - 121-7x + 242 = x1x - 123A1x - 122 + Cx1x - 12 + Dx4
-7x + 24 = A1x - 122 + Cx1x - 12 + Dx
We now work with equation (8). Let x = 0. Then
24 = A
Now let x = 1 in equation (8). Then
17 = D
Use A = 24 and D = 17 in equation (8) and collect like terms.
-7x + 24 = 241x Now let x = 2. Then
-14 + 24 =
-48 =
-24 =
122 + Cx1x - 12 + 17x
24 + C122 + 34
2C
C
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CHAPTER 10
Systems of Equations and Inequalities
We now know all the numbers A, B, C, D, and E, so, from equation (6), we have the
decomposition
x3 - 8
24
-24
17
-7
8
=
+
+
+ 2 +
2
3
2
x
x - 1
x 1x - 12
x
1x - 12
1x - 123
The method employed in Example 3, although somewhat tedious, is still
preferable to solving the system of five equations containing five variables that the
expansion of equation (6) leads to.
NOW WORK PROBLEM
19.
The final two cases involve irreducible quadratic factors. A quadratic factor is
irreducible if it cannot be factored into linear factors with real coefficients. A quadratic expression ax2 + bx + c is irreducible whenever b2 - 4ac 6 0. For example,
x2 + x + 1 and x2 + 4 are irreducible.
P
3 Decompose , where Q has a Nonrepeated Irreducible
✓
Q
Quadratic Factor
Case 3:
Q contains a nonrepeated irreducible quadratic factor.
If Q contains a nonrepeated irreducible quadratic factor of the form
P
ax2 + bx + c, then, in the partial fraction decomposition of , allow
Q
for the term
Ax + B
ax2 + bx + c
where the numbers A and B are to be determined.
EXAMPLE 4
Nonrepeated Irreducible Quadratic Factor
Write the partial fraction decomposition of
Solution
3x - 5
.
x3 - 1
We factor the denominator,
x3 - 1 = 1x - 121x2 + x + 12
and find that it has a nonrepeated linear factor x - 1 and a nonrepeated irreducible
A
quadratic factor x2 + x + 1. We allow for the term
by Case 1, and we allow
x
- 1
Bx + C
for the term 2
by Case 3. We write
x + x + 1
3x - 5
A
Bx + C
=
+ 2
x - 1
x3 - 1
x + x - 1
We clear fractions by multiplying each side
x3 - 1 = 1x - 121x2 + x + 12 to obtain the identity
(9)
of
equation
3x - 5 = A1x2 + x + 12 + 1Bx + C21x - 12
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(9)
by
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SECTION 10.5
Partial Fraction Decomposition
791
2
Now let x = 1. Then equation (10) gives -2 = A132, or A = - . We use this value
3
of A in equation (10) and simplify.
2
3x - 5 = - 1x2 + x + 12 + 1Bx + C21x - 12
3
313x - 52 = -21x2 + x + 12 + 31Bx + C21x - 12
Multiply each side by 3.
2
9x - 15 = -2x - 2x - 2 + 31Bx + C21x - 12
2x2 + 11x - 13 = 31Bx + C21x - 12
Collect terms.
12x + 1321x - 12 = 31Bx + C21x - 12
2x + 13 = 3Bx + 3C
2 = 3B and 13 = 3C
2
13
B =
C =
3
3
Factor the left side.
Cancel x - 1 on each side.
Equate coefficients.
From equation (9), we see that
2
2
13
x +
3x - 5
3
3
3
+ 2
=
3
x - 1
x - 1
x + x + 1
NOW WORK PROBLEM
21.
P
4 Decompose , where Q Has Repeated Irreducible Quadratic
✓
Q
Factors
Case 4:
Q contains repeated irreducible quadratic factors.
If the polynomial Q contains a repeated irreducible quadratic factor
1ax2 + bx + c2 , n Ú 2, n an integer, then, in the partial fraction decomposiP
tion of , allow for the terms
Q
n
A 1x + B1
2
+
A 2 x + B2
1ax + bx + c2
2
2
ax + bx + c
+ Á +
A nx + Bn
1ax2 + bx + c2
n
where the numbers A 1 , B1 , A 2 , B2 , Á , A n , Bn are to be determined.
EXAMPLE 5
Repeated Irreducible Quadratic Factor
Write the partial fraction decomposition of
Solution
x3 + x2
1x2 + 42
2
.
The denominator contains the repeated irreducible quadratic factor 1x2 + 42 , so
we write
x3 + x2
1x + 42
2
2
=
Ax + B
Cx + D
+
2
2
x + 4
1x2 + 42
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We clear fractions to obtain
x3 + x2 = 1Ax + B21x2 + 42 + Cx + D
Collecting like terms yields
x3 + x2 = Ax3 + Bx2 + 14A + C2x + D + 4B
Equating coefficients, we arrive at the system
A
B
d
4A + C
D + 4B
=
=
=
=
1
1
0
0
The solution is A = 1, B = 1, C = -4, D = -4. From equation (11),
x3 + x2
1x + 42
2
2
=
x + 1
-4x - 4
+
2
2
x + 4
1x2 + 42
NOW WORK PROBLEM
35.
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