College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
6
Systems of
Equations and
Inequalities
6.4
Partial Fractions
Introduction
To write a sum or difference of fractional
expressions as a single fraction, we bring
them to a common denominator.
1
1
(2 x  1)  ( x  1)
3x


 2
x  1 2x  1 ( x  1)(2 x  1)
2x  x  1
Introduction
However, for some applications of algebra
to calculus, we must reverse this process.
• We must express a fraction such as
3x/(2x2 – x – 1)
as the sum of the simpler fractions
1/(x – 1) and 1/(2x + 1)
Partial Fractions
These simpler fractions are called
partial fractions.
• In this section, we learn how to find them.
Partial Fractions
Let r be the rational function
P( x )
r (x) 
Q( x )
where the degree of P is less than
the degree of Q.
Partial Fractions
By the Linear and Quadratic Factors Theorem
in Section 4.5, every polynomial with real
coefficients can be factored completely into
linear and irreducible quadratic factors.
• That is, factors of the form
ax + b and ax2 + bx + c
where a, b, and c are real numbers.
Partial Fractions
For instance,
x4 – 1 = (x2 – 1)(x2 + 1)
= (x – 1)(x + 1)(x2 + 1)
Partial Fraction Decomposition
After we have completely factored
the denominator Q of r, we can express r(x)
as a sum of partial fractions of the form
A
i
(ax  b)
and
Ax  B
2
j
(ax  bx  c )
• This sum is called the partial fraction
decomposition of r.
Partial Fraction Decomposition
Let’s examine the details
of the four possible cases.
Case 1
The denominator is:
• A product of distinct linear factors.
Case 1
Suppose that we can factor Q(x) as
Q(x) = (a1x + b1 )(a2x + b2 ) ··· (anx + bn )
with no factor repeated.
• The partial fraction decomposition of P(x)/Q(x)
takes the form
A1
A2
An
P( x )


 ... 
Q( x ) a1x  b1 a2 x  b2
an x  bn
Partial Fraction Decomposition
The constants
A1, A2, . . . , An
are determined as in the following
example.
E.g. 1—Distinct Linear Factors
Find the partial fraction
decomposition of:
5x  7
3
2
x  2x  x  2
E.g. 1—Distinct Linear Factors
The denominator factors as:
x3 + 2x2 – x – 2 = x2(x + 2) – (x + 2)
= (x2 – 1)(x + 2)
= (x – 1)(x + 1)(x + 2)
• This gives the partial fraction
decomposition
5x  7
A
B
C



3
2
x  2x  x  2 x  1 x  1 x  2
E.g. 1—Distinct Linear Factors
Multiplying each side by the common
denominator, (x – 1)(x + 1)(x + 2), we get:
5x + 7
= A(x + 1)(x + 2) + B(x – 1)(x + 2)
+ C(x – 1)(x + 1)
= A(x2 + 3x + 2) + B(x2 + x – 2) + C(x2 – 1)
= (A + B + C)x2 + (3A + B)x + (2A – 2B – C)
E.g. 1—Distinct Linear Factors
If two polynomials are equal, their coefficients
are equal.
Thus, since 5x + 7 has no x2-term,
we have A + B + C = 0.
• Similarly, by comparing the coefficients of x,
we see that:
3A + B = 5
• By comparing constant terms, we get:
2A – 2B – C = 7
E.g. 1—Distinct Linear Factors
This leads to the following system of linear
equations for A, B, and C.
 A  B C 0

5
3 A  B
2 A  2B  C  7

Equation 1
Equation 2
Equation 3
• We use Gaussian elimination to solve this.
E.g. 1—Distinct Linear Factors
A  B  C  0

  2B  3C  5
  4B  3C  7

A  B  C  0

  2B  3C  5

3C  3

Equation 2 + (  3)  Equation 1
Equation 3 + (  2)  Equation 1
Equation 3 + (  2)  Equation 2
E.g. 1—Distinct Linear Factors
From the third equation, we get C = –1.
Back-substituting, we find that
B = –1 and A = 2
• So, the partial fraction decomposition is:
5x  7
2
1
1



3
2
x  2x  x  2 x  1 x  1 x  2
Partial Fraction Decomposition
The same approach works in
the remaining cases.
• We set up the partial fraction decomposition
with the unknown constants, A, B, C, . . . .
• Then, we multiply each side of the resulting
equation by the common denominator,
simplify the right-hand side of the equation,
and equate coefficients.
Partial Fractions
This gives a set of linear equations that
will always have a unique solution.
• This is provided that the partial fraction
decomposition has been set up correctly.
Case 2
The denominator is:
• A product of linear factors,
some of which are repeated.
Case 2
Suppose the complete factorization of Q(x)
contains the linear factor ax + b repeated k
times—that is, (ax + b)k is a factor of Q(x).
• Then, corresponding to each such factor,
the partial fraction decomposition for P(x)/Q(x)
contains
A1
A2
Ak

 ... 
2
ax  b (ax  b)
(ax  b)k
E.g. 2—Repeated Linear Factors
Find the partial fraction decomposition
of:
x 1
3
x( x  1)
2
• The factor x – 1 is repeated three times
in the denominator.
E.g. 2—Repeated Linear Factors
So, the partial fraction decomposition
has the form
x 1
A
B
C
D
 


3
2
3
x( x  1)
x x  1 ( x  1) ( x  1)
2
E.g. 2—Repeated Linear Factors
We then multiply each side by the common
denominator x(x – 1)3.
x2 + 1
= A(x – 1)3 + Bx(x – 1)2 + Cx(x – 1) + Dx
= A(x3 – 3x2 + 3x – 1) + B(x3 – 2x2 + x)
+ C(x2 – x) + Dx
= (A + B)x3 + (–3A – 2B +C)x2
+ (3A + B – C + D)x – A
E.g. 2—Repeated Linear Factors
Equating coefficients, we get:
0
 A B
3 A  2B  C
1


 3A  B  C  D  0
  A
1
• If we rearrange these by putting the last one
in the first position, we can easily see (using
substitution) that the solution to the system is:
A = –1, B = 1, C = 0, D = 2
E.g. 2—Repeated Linear Factors
So, the partial fraction decomposition
is:
x 1
1
1
2



3
3
x( x  1)
x x  1 ( x  1)
2
Case 3
The denominator has:
• Irreducible quadratic factors,
none of which is repeated.
Case 3
Suppose the complete factorization of Q(x)
contains the quadratic factor ax2 + bx + c
(which can’t be factored further).
• Then, corresponding to this, the partial fraction
decomposition of P(x)/Q(x) will have a term
of the form
Ax  B
2
ax  bx  c
E.g. 3—Distinct Quadratic Factors
Find the partial fraction decomposition
of:
2
2x  x  4
3
x  4x
• Since x3 + 4x = x(x2 + 4), which can’t be
factored further, we write:
2x  x  4 A Bx  C
  2
3
x  4x
x x 4
2
E.g. 3—Distinct Quadratic Factors
Multiplying by x(x2 + 4), we get:
2x2 – x + 4 = A(x2 + 4) + (Bx + C)x
= (A + B)x2 + Cx + 4A
E.g. 3—Distinct Quadratic Factors
Equating coefficients gives us:
A  B  2

 C  1
 4A  4

• So, A = 1, B = 1, and C = –1.
E.g. 3—Distinct Quadratic Factors
The required partial fraction
decomposition is:
2x  x  4 1 x  1
  2
3
x  4x
x x 4
2
Case 4
The denominator has:
• A repeated irreducible quadratic
factor.
Case 4
Suppose the complete factorization of Q(x)
contains the factor (ax2 + bx + c)k, where
ax2 + bx + c can’t be factored further.
• Then, the partial fraction decomposition
of P(x)/Q(x) will have the terms
A1x  B1
A2 x  B2

2
ax  bx  c
ax 2  bx  c


2
 ... 
Ak x  Bk
ax
2
 bx  c

k
E.g. 4—Repeated Quadratic Factors
Write the form of the partial fraction
decomposition of:
x  3 x  12 x  1
5
x
3
x
2
2

 x 1 x  2
2

3
E.g. 4—Repeated Quadratic Factors
x  3 x  12 x  1
5
x
3
x
2
2

 x 1 x  2
2

3
A B C
Dx  E
Fx  G
  2 3 2
 2
x x
x
x  x 1 x  2
Hx  I
Jx  K


2
3
2
2
x 2
x 2

 

Repeated Quadratic Factors
To find the values of
A, B, C, D, E, F, G, H, I, J, and K
in Example 4, we would have to solve
a system of 11 linear equations.
• Although possible, this would certainly involve
a great deal of work!
Using Long Division
The techniques we have described in
this section apply only to:
• Rational functions P(x)/Q(x) in which
the degree of P is less than the degree of Q.
If this isn’t the case, we must first use
long division to divide Q into P.
E.g. 5—Long Division to Prepare for Partial Fractions
Find the partial fraction decomposition
of:
2x  4 x  2x  x  7
3
2
x  2x  x  2
4
3
2
• The degree of the numerator is larger
than that of the denominator.
E.g. 5—Long Division to Prepare for Partial Fractions
So, we use long division.
2x
x 3  2x 2  x  2 2x 4  4 x 3  2x 2  x  7
2x  4 x  2x  4 x
4
3
2
5x  7
E.g. 5—Long Division to Prepare for Partial Fractions
Thus, we obtain:
2x  4 x  2x  x  7
3
2
x  2x  x  2
5x  7
 2x  3
2
x  2x  x  2
4
3
2
• The remainder term now satisfies the requirement
that the degree of the numerator is less than that
of the denominator.
E.g. 5—Long Division to Prepare for Partial Fractions
At this point, we proceed as in
Example 1 to obtain the decomposition
2x  4 x  2x  x  7
3
2
x  2x  x  2
2
1
1
 2x 


x 1 x 1 x  2
4
3
2