section 6.3 partial fractions

Section 6.3 Partial Fractions

Advanced Algebra

What is Partial Fraction Decomposition?

There are times when we are working with Rational Functions of the form when we want to

split it up into two simpler fractions. The process that we go through to do this is called “partial fraction decomposition”

)()(xgxf

Let’s say you had the following:

14

43

xx

)1)(4(16433

xxxx

)4()4(

)1(4

)1()1(

)4(3

xx

xxx

x

What would you need to do to add them together?

Multiply by a common denominator to both fractions

Write as one fraction and simplify the numerator

)1)(4(137

xx

xThe final answer

Partial Fraction Decomposition is the reverse of what we just did here…

22xx

222

xB

xA

xx

2

22222

xxBx

xxAx

xxxx

BxxA 22

BxAAx 22

A22

A1

BxAx 0

BA0B10

B 1 211

xx

Cancel anything necessary

Multiply by LCD

Break into 2 smaller fractions

Collect like terms and set equal

5 33 1x

x x

3 1A Bx x

Multiply by the common denominator.

5 3 1 3x A x B x

5 3 3x Ax A Bx B Collect like terms together and set them equal to each other.

5x Ax Bx 3 3A B

5 A B 3 3A B Solve two equations with two unknowns.

3235

2 xx

x

5 33 1x

x x

3 1A Bx x

5 3 1 3x A x B x

5 3 3x Ax A Bx B

5x Ax Bx 3 3A B

5 A B 3 3A B Solve two equations with two unknowns.

5 A B 3 3A B

3 3A B

8 4B

2 B 5 2A

3 A

This technique is calledPartial Fractions

12

33

xx

3235

2 xx

x

1310211

2 xx

x

12151310211

2

xB

xA

xxx

)15()12(211 xBxAx

BxBAxAx 1512211

xBxAx 5211 BA 2

BA 5211

BA 5211 BA 224

B77

B1

12 A

A 3

A3

121

153

xx

2

6 72

xx

Repeated roots: we must use two terms for partial fractions.

22 2A Bx x

6 7 2x A x B

6 7 2x Ax A B

6x Ax 7 2A B

6 A 7 2 6 B

7 12 B

5 B

2

6 52 2x x

Sometimes you might get a repeated factor (multiplicity) in the denominator

Partial-Fraction Decomposition Repeated linear factor

2

2

12374

xxxx

22

2

11212374

xC

xB

xA

xxxx

2121374 22 xCxxBxAxx

2212374 222 xCxxBxxAxx

CCxBBxBxAAxAxxx 222374 222

2224 BxAxx CxBxAxx 27 CBA 223

CxBxAxx 27 CBA 223 2224 BxAxx

BA4 CBA 27 CBA 22414

BA 4511

BA 4511

BA 4416

A927 A3

B34B1

C21233

C213

C24

C 2

212

11

23

xxx

Now we have 3 equations with 3 unknowns. Solve like in previous section.

3 2

2

2 4 32 3

x x xx x

If the degree of the numerator is higher than the degree of the denominator, use long division first.

2 3 22 3 2 4 3x x x x x 2x

3 22 4 6x x x 5 3x

2

5 322 3xx

x x

5 323 1xx

x x

3 2 23 1

xx x

(from example one)

What if the denominator has a non-factorable quadratic in it?

21)2)(1(47

22

2

xC

xBAx

xxxx

0242

7 222

CBxBxAx

xCxAx

)1()2)((047 22 xCxBAxxx

Partial Fraction Decomposition can get very complicated, very quickly when there are non-factorable quadratics and repeated linear factors…here is an easy example…

CCxBBxAxAxxx 222 22047

)2)(1(47

2

2

xxxx

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)…

)1)(5(2

xx

x

15)1)(5(2

xB

xA

xxx A

43

1525

B

41

5121

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…

Do the same for the other linear factor

141

543

)1)(5(2

xxxxx Which should probably

be simplified…

22

2 41 1x

x x

irreduciblequadratic

factor

repeated root

22 1 1 1Ax B C Dx x x

first degree numerator

2 2 22 4 1 1 1 1x Ax B x C x x D x

2 3 2 22 4 2 1 1x Ax B x x C x x x Dx D

3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D

A challenging example:

3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D

0 A C 0 2A B C D 2 2A B C 4 B C D

1 0 1 0 02 1 1 1 0

1 2 1 0 20 1 1 1 4

2 r 3

r 1

1 0 1 0 00 3 1 1 40 2 0 0 20 1 1 1 4

2

1 0 1 0 00 1 0 0 10 3 1 1 40 1 1 1 4

3 r 2

r 2

1 0 1 0 00 1 0 0 10 0 1 1 10 0 1 1 3

r 3

1 0 1 0 00 1 0 0 10 3 1 1 40 1 1 1 4

3 r 2

r 2

1 0 1 0 00 1 0 0 10 0 1 1 10 0 1 1 3

r 3

1 0 1 0 00 1 0 0 10 0 1 1 10 0 0 2 2

2

1 0 1 0 00 1 0 0 10 0 1 1 10 0 0 1 1

r 4

1 0 1 0 00 1 0 0 10 0 1 0 20 0 0 1 1

r 3

1 0 0 0 20 1 0 0 10 0 1 0 20 0 0 1 1

1 0 0 0 20 1 0 0 10 0 1 0 20 0 0 1 1

22

2 41 1x

x x

22 1 1 1Ax B C Dx x x

22

2 1 2 11 1 1

xx x x

We can do this problem on the TI-89:

22

2 4expand1 1x

x x

expand ((-2x+4)/((x^2+1)*(x-1)^2))

22 2

2 1 2 11 1 1 1x

x x x x

Of course with the TI-89, we could just integrate and wouldn’t need partial fractions!

3F2