Steps for Partial Fraction Decomposition
1.
Determine the degree of the numerator and the degree of the denominator.
Deg Num = N and Deg Den = D
If the N is larger than or equal to D, then divide (numerator)
(denominator) and continue the steps with the remainder fraction.
Num .
÷
Den .
=
Quotient
+
Re mainder
Divisor
2.
Factor the denominator into prime factors which are linear ( ax
+ b ) , quadratic ( ax
2 + bx
+ c ) or higher polynomials.
3.
Set-up step. This steps results in an equation.
Use the factors in the denominator to set-up the partial fraction decomposition. Each factor in the denominator will yield some number of partial fractions. The exponent or power of the factor will determine how many partial fractions come from that particular factor.
Linear factors:
( ax
+ b ) n
--- from this factor there will be “ n ” partial fractions of the form
A
( ax
+ b
+
1
) (
B ax
+ b
+
2
) (
C ax
+ b )
3
H
( ax
+ b ) n
Notice the numerators of these fractions. If the set of parenthesis in the denominator contain a
LINEAR expression, then you place a CONSTANT in the numerator of that fraction.
Quadratic factors:
( ax
2 + bx
+ c ) m
--- from this factor there will be “ m ” partial fractions of the form
Ax
+
B
( ax
2 + bx
+ c
+
1
) (
Cx
+
D ax
2 + bx
+ c
+
2
) (
Ex
+
F ax
2 + bx
+ c )
3
( ax
2
Jx
+
K
+ bx
+ c ) m
Notice the numerators of these fractions. If the set of parenthesis in the denominator contain a
QUADRATIC expression, then you place a LINEAR expression in the numerator of that fraction.
4.
Multiply both sides of the equation by the LCD (the original denominator in factored form).
5.
Solve for A, B, C, etc. using Method 1 or Method 2.
Method 1 – substitute chosen values of x into the equation
Method 2 – - multiply out the right-hand side of the equation
- rearrange the rhs in descending order
- equate coefficients of powers of x from each side of the equation
- Use matrices to solve the resulting system of equations
6.
Write out the partial fraction decomposition and state your answer clearly.

Partial Fractions example of “Method 2”
Example: Find the partial fraction decomposition for
3 x
2 −
3 x
−
8
( x
−
5)( x
2 + −
4)
Step 1. Degree Num: 2 Degree Den: 3
Æ
go to step 2
Step 2. Factor the denominator into prime factors.
Step 3. Set-up step
3 x
2 −
3 x
−
8
( x
−
5)( x
2 + −
4)
=
A
( x
−
5)
+
( x
2
Bx
+
C
+ −
4)
Step 4. Multiply through by the factored denominator.
( x
−
5)( x 2 + −
4)
3 x
2 −
3 x
−
8
( x
−
5)( x
2 + −
4)
=
3 x
2 −
3 x
−
8
=
A
( x
−
5)
( x
−
5)( x 2 + −
4)
+
( x
2
Bx
+
C
+ −
4)
( x
−
5)( x 2 + −
4)
A x
2 x 4)
+
( Bx
+
)(
−
5)
Step 5. Solve for A, B, C using Method 2. Multiply out the right hand side of the equation. Then rearrange and regroup so that you will have an “
2 x ” term, an “ x ” term, and a constant term.
3
3
3
3 x x x x
3 x
2
2
2
2
2
−
−
−
−
3
3
3
3 x x x x
−
−
−
−
8
−
3 x
−
8
8
8
8
=
=
=
=
=
A x
Ax
A x
2
2
2
+
+ x
Ax
A x
2 +
B x
( A
+
B ) x
2
2
−
+
+
4)
4 A
+
+
A x
−
4 A
+
(
Bx
Bx
B x
2
2 −
+
5
)(
Bx
+
−
5)
Cx
5 B x
+
C x
−
5 C
−
5 C
A x
5 B x
+
C x
−
4 A
−
5 C
( A
5 B
C ) x
−
4 A
−
5 C rearran x
2
ter m x term constan t ter m ge terms
Now equate the coefficients of
2 x , the coefficients of x , and the constant terms on each side.
3 x
2 −
3 x
−
8
=
( A
+
B ) x
2 +
( A
−
5 B
+
C ) x
−
4 A
−
5 C x
2 coefficients: A
+
B x coefficient s: A
−
5 B
+
C constant coeffic ien t s: -4 A
−
5 C
=
3
= −
3
= −
8
Now use matrices to solve this system of 3 equations and 3 unknowns. rref
⎛
⎜
⎝
⎢
⎢
⎡
⎢
⎣
−
1 1 0 3
1
−
5
−
1
−
−
3
⎤
⎥
⎥
⎦
⎞
⎟
⎟
⎠
=
⎡
⎢
⎢
⎣
1 0 0 2
0 1 0 1
4 0 5 8
⎥ ⎢
0 0 1 0
⎥
⎦
⎤
⎥
So, A
=
2 , B
=
1, C
=
0 and the partial fraction decomposition is
3 x
2 −
3 x
−
8
( x
−
5)( x
2 + −
4)
=
=
2
−
+
( x 5) (
1 x
+
0 x
2 + −
4)
A
( x
−
5)
+
Bx
+
C
( x
2 + −
4)
=
2
−
+
( x 5) ( x x
2 + −
4)