1
Section 12.5 – Partial Function Decomposition
If we were asked to add two rational expressions like
3
x− 2
+
2x+ 5
x2 +4
, we
would find the common denominator, build the fractions, and then combine
the numerators:
3
x− 2
+
2x+ 5
x2 +4
=
3(x2 +4)
(x −2)(x 2 +4)
+
(2x +5)(x−2)
(x 2 +4)(x−2)
=
3x2 +12+2x 2 +x −10
(x −2)(x2 +4)
In this section, we want to go backwards. Given
3
x− 2
break it down to
+
2x+ 5
x2 +4
5x2 + x+2
(x −2)(x 2 +4)
=
5x2 + x+2
(x −2)(x 2 +4)
.
, we want to
. This procedure is called partial fraction
decomposition.
Objective 1 – 4
Partial Fraction Decomposition
Partial Fraction Decomposition
Let R(x) be a rational expression reduced to lowest terms equal to
P(x)
Q(x)
where the degree of P(x) < the degree of Q(x). (If not, divide and work with
the remainder).
1)
Let x – a be a linear factor of Q(x). If (x – a)n is the highest power of
x – a that divides Q(x), then we will have the fractions:
A1
x− a
+
A2
(x −a)2
+…+
An
(x −a)n
as part of the decomposition where A1, …, An are to be determined.
We will need to do this for each distinct linear factor of Q(x).
2)
€
Let ax2 + bx + c be a quadratic factor of Q(x) that cannot be factored
into a product of linear factors with real coefficients (irreducible). If
(ax2 + bx + c)m is the highest power of ax2 + bx + c that divides Q(x),
then we will have the fractions:
A 1x +B1
2
ax +bx+c
+
A 2 x+ B2
2
(ax +bx +c)
2
+…+
A m x+Bm
(ax2 +bx +c) m
as part of the decomposition where A1, …, Am and B1, …, Bm are to
be determined. We will need to do this for each distinct irreducible
quadratic factor of Q(x).
€
3)
Set the original function equal to the sum of all the partial functions
and multiply both sides by the L.C.D. Then, solve for the coefficients.
2
Write the partial function decomposition of the following:
Ex. 1
R(x) =
4x+ 13
x2 − x− 2
Solution:
First, factor the denominator:
4x +13
4x+ 13
=
The denominator has two distinct linear
2
(x −2)(x +1)
x − x− 2
factors of degree 1. Thus, we need one partial fraction for each one:
4x +13
A
B
=
+
(Multiply both sides by (x – 2)(x + 1))
(x −2)(x +1)
x− 2
x+ 1
4x + 13 = A(x + 1) + B(x – 2)
If we let x = – 1, we will get:
4(– 1) + 13 = A((– 1) + 1) + B((– 1) – 2)
– 4 + 13 = 0 – 3B
9 = – 3B or B = – 3
If we let x = 2, we will get:
4(2) + 13 = A((2) + 1) + B((2) – 2)
8 + 13 = 3A + 0
21 = 3A or A = 7.
4x +13
7
−3
7
3
Hence,
=
+
or
–
(x −2)(x +1)
Ex. 2
R(x) =
x− 2
x+ 1
x− 2
x+ 1
− x+ 14
x3 −8
Solution:
First, factor the denominator:
− x+ 14
− x+ 14
=
The denominator has a linear factor of
3
2
x −8
(x −2)(x +2x+ 4)
degree 1 and an irreducible quadratic factor of degree 1. Thus, we
need one partial fraction for each one:
A
− x+ 14
Bx +C
=
+ 2
(Multiply both sides by the L.C.D.)
2
x− 2
(x −2)(x +2x+ 4)
2
x +2x+ 4
– x + 14 = A(x + 2x + 4) + (Bx + C)(x – 2)
If we let x = 2, we will get:
– (2) + 14 = A((2)2 + 2(2) + 4) + (B(2) + C)((2) – 2)
– 2 + 14 = 12A + 0
12 = 12A or A = 1
Replace A by 1 in – x + 14 = A(x2 + 2x + 4) + (Bx + C)(x – 2):
– x + 14 = x2 + 2x + 4 + (Bx + C)(x – 2)
If we let x = 0, we will get:
3
– (0) + 14 = (0)2 + 2(0) + 4 + (B(0) + C)((0) – 2)
0 + 14 = 0 + 0 + 4 + – 2C
14 = 4 – 2C
10 = – 2C or C = – 5
Replace C by – 5 in – x + 14 = x2 + 2x + 4 + (Bx + C)(x – 2):
– x + 14 = x2 + 2x + 4 + (Bx – 5)(x – 2)
If we let x = 1, we will get:
– (1) + 14 = (1)2 + 2(1) + 4 + (B(1) – 5)((1) – 2)
– 1 + 14 = 1 + 2 + 4 – B + 5
13 = 12 – B
1 = – B or B = – 1
1
1
− x+ 14
−x −5
x+ 5
So,
=
+
or
–
2
2
2
(x −2)(x +2x+ 4)
x− 2
x− 2
x +2x+ 4
x +2x+ 4
5x −75
Ex. 3
3
x +7x 2 +15x +9
Solution:
First, factor the denominator:
5x −75
We will need to use synthetic division.
3
2
x +7x +15x +9
–1 | 1
|
1
Thus,
5x −75
3
7
–1
6
2
x +7x +15x +9
15
–6
9
=
9
–9
|0
5x− 75
2
(x +1)(x +6x+ 9)
=
5x− 75
(x +1)(x +3)2
. The denominator
has a linear factor of degree 1 and linear factor of degree 2. Thus, we
need one partial fraction for first one and two for the second:
A
B
5x− 75
C
=
+
+
(Multiply both sides by the L.C.D.)
2
2
(x +1)(x +3)
x+ 1
x+ 3
(x +3)
2
5x – 75 = A(x+ 3) + B(x + 1)(x + 3) + C(x + 1)
If we let x = – 3, we will get:
5(– 3) – 75 = A((– 3)+ 3)2 + B((– 3) + 1)((– 3) + 3) + C((– 3) + 1)
– 15 – 75 = A(0) + B(0) + C(– 2)
– 90 = – 2C or C = 45
If we let x = – 1, we will get:
5(– 1) – 75 = A((– 1)+ 3)2 + B((– 1) + 1)((– 1) + 3) + C((– 1) + 1)
– 5 – 75 = A(4) + B(0) + C(0)
– 80 = 4A or A = – 20
Plug A and C into 5x – 75 = A(x+ 3)2 + B(x + 1)(x + 3) + C(x + 1):
5x – 75 = – 20(x+ 3)2 + B(x + 1)(x + 3) + 45(x + 1)
4
If we let x = 0, we will get:
5(0) – 75 = – 20((0)+ 3)2 + B((0) + 1)((0) + 3) + 45((0) + 1)
– 75 = – 20(9) + B(3) + 45(1)
– 75 = – 180 + 3B + 45
– 75 = – 135 + 3B
60 = 3B or B = 20
5x− 75
Hence,
(x +1)(x +3)
2
−20
x +1
=
+
20
x+ 3
+
45
(x +3)2
4x3 −3x2
Ex. 4
x 4 +18x2 +81
Solution:
First, factor the denominator:
4x3 −3x2
x 4 +18x2 +81
4x3 −3x2
=
(x2 +9) 2
The denominator has an irreducible factor of
degree 2. Thus, we need two partial fractions:
4x3 −3x2
2
(x +9)
2
=
Ax+ B
2
x +9
+
Cx +D
(x 2 +9)2
(Multiply both sides by the L.C.D.)
4x3 – 3x2 = (Ax + B)(x2 + 9) + Cx + D
Instead of plugging in numbers, we will expand and match up the
terms:
4x3 – 3x2 = (Ax + B)(x2 + 9) + Cx + D
4x3 – 3x2 = Ax3 + 9Ax + Bx2 + 9B + Cx + D (Group like terms)
4x3 – 3x2 = Ax3 + Bx2 + 9Ax + Cx + 9B + D
4x3 – 3x2 = Ax3 + Bx2 + (9A + C)x + (9B + D)
4x3 – 3x2 + 0x + 0 = Ax3 + Bx2 + (9A + C)x + (9B + D)
x3-term:
4x3 = Ax3 or A = 4
x2-term:
– 3x2 = Bx2 or B = – 3
x-term:
0x = (9A + C)x or 0 = 9A + C
constant: 0 = 9B + D
Since A = 4, then
Since B = – 3, then
0 = 9A + C = 9(4) + C
0 = 9B + D = 9(– 3) + D
0 = 36 + C o or C = – 36
0 = – 27 + D or D = 27
So,
4x3 −3x2
2
(x +9)
2
=
(4)x+ (− 3)
2
x +9
+
(− 36)x+(27)
2
2
(x +9)
=
4x− 3
2
x +9
+
−36x+ 27
(x 2 +9)2
.