7.8 Partial Fractions

If we wanted to add something like this:

2

2 5

3 12 5

3 12( 1) 5( 3) 2 2 5 15 7 13

( 1)( 3) ( 1)( 3)

1 3

1 3

2 3

x x

x xx x x x x

x x

x

x x x

x

x

x

x

We would need to find a common denominator!

Now, we want to REVERSE the process!

Go from here, to here!

This process is called partial fraction decomposition.

Steps: (1) Factor the denominator

(2) Express as two (or more) fractions with A & B (or more) as placeholders for the numerators

(3) Multiply both sides by LCD

(4) Solve for A & B (or more) by letting x equal values that would make the other letter be multiplied by 0

(5) Substitute A & B

(sometimes simple, sometimes not!)

2

2 8 ( 1) ( 3)

let 1: let 3 :

2(1) 8 (1 1) (1 3) 2(3) 8 (3 1) (3 3)

2 8 (0) ( 2) 6 8 (2) (0)

6 2 2 2

3 1

2 8 1 3

4 3 3 1

x A x B x

x x

A B A B

A B A B

B A

B A

x

x x x x

For all examples, decompose into partial fractions.

Ex 1) 2

2 8 2 8

4 3 ( 3)( 1) 3 1

x x A B

x x x x x x

( 3)( 1)x x

2

2

3 2

6 11 15 ( 5)( 3) ( )( 3) ( )( 5)

let 0 : let 5 : let 3 :

15 ( 15) 80 (40) 72 (24)

1 2 3

6 11 15 1 2 3

2 15 5 3

x x A x x B x x C x x

x x x

A B C

A B C

x x

x x x x x x

Ex 2) 2 2

3 2

6 11 15 6 11 15

2 15 ( 5)( 3) 5 3

x x x x A B C

x x x x x x x x x

( 5)( 3)x x x

x(x2 – 2x – 15)

2 2

2

2 14 15 ( 2) ( 1)( 2) ( 1)

let 2 : let 1:

5 (1) 3 (1)

5 3

15 3(0 2) (0 1)(0 2) 5(0 1)

15 12 2 5

2 2 1

x x A x B x x C x

x x

C A

C A

B

B

B B

If (x – a)n is a factor of the denominator, you will need (x – a), (x – a)2, …, (x – a)n

Ex 3) 2

2 2

2 14 15

( 1)( 2) 1 2 ( 2)

x x A B C

x x x x x

2( 1)( 2)x x

To solve for B, use A & C & let x = 0:

2

3 1 5

1 2 ( 2)x x x

Ex 4)

Factor bottom!

1 6 1 –10 3↓

66 77

–3–3 0

6x2 + 7x – 3 6x2 + 9x – 2x – 33x(2x + 3) – 1(2x + 3) (3x – 1)(2x + 3)

denom = (x – 1)(3x – 1)(2x + 3)

2

3 2

5 16 1

6 10 3

x x

x x x

–18

7

–29

2

32

9 3 5 114 2 2 2

55 554 4

13

1 1 2 119 3 3 3

44 229 9

5 16 1 (2 3)(3 1) ( 1)(3 1) ( 1)(2 3)

let 1: let :

5 16 1 (5)(2) 5( ) 16( ) 1 ( )( )

20 10

2 1

let :

5( ) 16( ) 1 ( )( )

2

x x A x x B x x C x x

x x

A B

A B

A B

x

C

C

C

Ex 4) cont… ( 1)(2 3)(3 1)x x x

25 16 1

( 1)(2 3)(3 1) 1 2 3 3 1

x x A B C

x x x x x x

2 1 2

1 2 3 3 1x x x

Homework

#708 Pg 378 #1, 5, 7, 18, 21, 29