```Section 6.3 Partial Fractions
What is Partial Fraction
Decomposition?
There are times when we are working with Rational
f ( x)
Functions of the form
when we want to
g ( x)
split it up into two simpler fractions. The process
that we go through to do this is called “partial
fraction decomposition”
Let’s say you had the following:
What would you need to do
to add them together?
3 ( x  1)
4 ( x  4)

( x  4) ( x  1) ( x  1) ( x  4)
3x  3  4 x  16
( x  4)( x  1)
7 x  13
( x  4)( x  1)
3
4

x  4 x 1
Multiply by a common
denominator to both
fractions
Write as one fraction and simplify
the numerator
Partial Fraction Decomposition is the reverse of what we
just did here…

2
x x  2 
2
A
B
 
x x  2  x x  2
Break into 2 smaller
fractions
2 xx  2 Axx  2 Bxx  2


x x  2 
x
x2
2  Ax  2  Bx
Multiply by LCD
Cancel anything
necessary
2  Ax  2 A  Bx
2  2A
1 A
0  Ax  Bx
0  A B
Collect like terms and set equal
0 1 B
1  B
1
1

x x2
5x  3
x2  2x  3
5x  3
A
B


 x  3 x  1 x  3 x  1
Multiply by the common
denominator.
5x  3  A x 1  B  x  3
5 x  3  Ax  A  Bx  B  3
5x  Ax  Bx
5  A B
3  A  B  3
3  A  3B
Collect like terms
together and set them
equal to each other.
Solve two equations with
two unknowns.

5  A B
5x  3
x2  2x  3
3  A  3B
3   A  3B
8  4B
5x  3
A
B


 x  3 x  1 x  3 x  1
2B
5  A2
3 A
5x  3  A x 1  B  x  3
5 x  3  Ax  A  Bx  B  3
5x  Ax  Bx
5  A B
3  A  B  3
3  A  3B
3
2

x  3 x 1
This
is called
Solvetechnique
two equations
with
Fractions
twoPartial
unknowns.

11  2 A  5B
 4  2 A  2 B
11x  2
10 x 2  3x  1
11x  2
A
B


2
10 x  3x  1 5 x  1 2 x  1
1 B
11x  2  A(2 x  1)  B(5x  1)
 2   A 1
11x  2  A2 x  1A  B5 x  1B
11x  A2 x  B5 x
 2  A  B
7  7B
3  A
3 A
11  2 A  5B
3
1

5x  1 2x 1
Sometimes you might get a repeated factor
(multiplicity) in the denominator
6x  7
A
B
Repeated roots: we must


2
2
use two terms for partial
x

2
 x  2
 x  2
fractions.
6 x  7  A  x  2  B
6 x  7  Ax  2 A  B
6x  Ax
7  2A  B
6 A
7  26  B
7  12  B
5  B
6
5

x  2  x  2 2

Partial-Fraction Decomposition Repeated linear factor
4x 2  7x  3
x  2x  12
4x 2  7 x  3
A
B
C



2
2
x

2
x

1
x  2x 1
x 1
4x2  7 x  3  Ax 1  Bx  2x 1  Cx  2
2

 

4x 2  7 x  3  A x 2  2x  1  B x 2  x  2  Cx  2
4 x 2  7 x  3  Ax2  2 Ax  A  Bx2  Bx  2B  Cx  2C
4 x 2  Ax2  Bx2
 7 x  2 Ax  Bx  Cx
 3  A  2 B  2C
4 x 2  Ax2  Bx2
 7 x  2 Ax  Bx  Cx
4  A B
 7  2 A  B  C
Now we have 3 equations with 3
unknowns. Solve like in previous section.
11  5 A  4 B
4  3 B
16  4 A  4 B
1 B
27  9 A
3 A
 3  A  2 B  2C
14  4 A  2 B  2C
11  5 A  4 B
 3  3  21  2C
 3  1  2C
 4  2C
3
1
2


x  2 x  1 x  12
2C
2 x3  4 x 2  x  3
x2  2 x  3
If the degree of the numerator is
higher than the degree of the
denominator, use long division first.
2x
x 2  2 x  3 2 x3  4 x 2  x  3
2 x3  4 x 2  6 x
5x  3
5x  3
2x  2
x  2x  3
(from example one)
5x  3
3
2
2x 
 2x 

 x  3 x  1
 x  3  x  1

What if the denominator has a nonfactorable quadratic in it?
7 x  4x
2
( x  1)( x  2)
2
Partial Fraction Decomposition can get very complicated,
very quickly when there are non-factorable quadratics and
repeated linear factors…here is an easy example…
7 x2  4x
Ax  B
C
 2

2
( x  1)( x  2) x  1 x  2
7 x  4 x  0  ( Ax  B)( x  2)  C ( x  1)
2
2
7x2  4x  0  Ax2  2 Ax  Bx  2B  Cx 2  C
Ax2  Cx 2  7 x 2
 2 Ax  Bx  4 x
 2B  C  0
Now just solve the 3 by 3 system…
A=3, B=2 and C=4
A nice shortcut if you have non-repeated linear factors—the
Heaviside Shortcut—named after mathematician Oliver
Heaviside (1850-1925)…
x2
( x  5)( x  1)
Tell yourself, “if x is 5, then x-5 is 0.”
Cover up the x-5 and put 5 in for the x
in what is left…
x2
A
B


( x  5)( x  1) x  5 x  1
Do the same for the other linear factor
3
1
x2
4
4


( x  5)( x  1) x  5 x  1
52 3
 A
5 1 4
1 2 1
 B
1 5 4
Which should probably
be simplified…
A challenging
example:
x
first degree numerator
2 x  4
2

 1  x  1
irreducible
factor
2
Ax  B
C
D
 2


x  1 x  1  x  12
repeated root




2 x  4   Ax  B  x  1  C x 2  1  x  1  D x 2  1
2




2 x  4   Ax  B  x 2  2 x  1  C x 3  x 2  x  1  Dx 2  D
2 x  4  Ax3  2 Ax2  Ax  Bx2  2Bx  B  Cx3  Cx 2  Cx  C  Dx 2  D

2 x  4  Ax3  2 Ax2  Ax  Bx2  2Bx  B  Cx3  Cx 2  Cx  C  Dx 2  D
0  A  C 0  2A  B  C  D
1
0
1
2  A  2 B  C
4  B C  D
0
0
1
0
1
0
0
2 1 1
1 2 1
1
0
0 2  r 3
2  r 1
0
0
1
3
0
1
0
1
1
4 3  r 2
0
1
1
1
4
0
1
1
1
4
1
0
1
0
0
1
0
1
0
0
0
0
3
2
1
0
1
0
4
2   2
0
0
1
0
0
1
0
1
1
1
0
1
1
1
4
0
0
1
1
3
r2
r3

1
0
1
0
0
1
0
1
0
0
0
0
1
3
0
1
0
1
1
4 3  r 2
0
0
1
0
0
1
0
1
1
1  r 4
0
1
1
1
4
0
0
0
1
1
1
0
1
0
0
1
0
1
0
0
0
0
1
0
0
1
0
1
1
1
0
0
1
0
0
1
0
0
1
2
0
0
1
1
3
0
0
0
1
1
1
0
1
0
0
1
0
0
0
2
0
0
1
0
0
1
0
1
1
1
0
0
1
0
0
1
0
0
1
2
0
0
0
2
2
0
0
0
1
1
r2
r3
 2
r3

x
2 x  4
2

 1  x  1
2
Ax  B
C
D
 2


2
x  1 x  1  x  1
2x 1
2
1
 2


x  1 x  1  x  12
We can do this problem on the TI-89:


2  x  4

expand  2
2
  x  1   x  1 


expand ((-2x+4)/((x^2+1)*(x-1)^2))
1 0 0 0 2
F2
03 1 0 0 1
0 0 1 0 2
2 x
1
2
1
Of course with the TI-89, we could



0 integrate
0 0and wouldn’t
1 1 need
just
x2  1 x2  1 x  1  x  12
partial fractions!
p
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