PAR TIALFRAC TION

+

DECOMPOSITION

2

3

3

4

xx

Let’s add the two fractions below. We need a common denominator:

2

2

x

x

3

3

x

x

32

9384

xx

xx

6

172

xx

x In this section we are going to learn how to take this answer and “decompose” it meaning break down into the fractions that were added together to get this answer

6

172

xx

x We start by factoring the denominator.

23

17

xx

x

There could have been a fraction for each factor of the denominator but we don’t know the numerators so we’ll call them A and B.

23

x

B

x

A

Now we’ll clear this equation of fractions by multiplying every term by the common denominator.

(x+3)(x-2)

(x+3)(x-

2)(x+

3)(x-

2)

3217 xBxAx

This equation needs to be true for any value of x.

We pick an x that will “conveniently” get rid of one of the variables and solve for the other.

3217 xBxAx

“The Convenient x Method for Solving”

Let x = -3

3323137 BAB(0) = 0

A520 4A

6

172

xx

x

Now we’ll “conveniently” choose x to be 2 to get rid of A and find B.

3217 xBxAxLet x = 2

3222127 BAA(0) = 0

B515 3B

23

17

xx

x

23

x

B

x

A

4A

4 3

6

172

xx

x

Summary of Partial Fraction Decomposition When Denominator Factors Into Linear Factors

(Factors of first degree)

Factor the denominator

Set fraction equal to sum of fractions with each factor as a denominator using A, B, etc.

for numerators

Clear equation of fractions

Use “convenient” x method to find A, B, etc.

Next we’ll look at repeated factors and quadratic factors

Partial Fraction Decomposition With Repeated Linear Factors

2

2

21

2

xx

x

When the denominator has a repeated linear factor, you need a fraction with a denominator for each power of the factor.

2221

x

C

x

B

x

A(x-1)(x

+2)2

(x-1)(x

+2)2

(x-1)(x

+2)2

(x-1)(x

+2)2

12122 22 xCxxBxAx

Let x = 1 1121112121 22 CBA

A93 3

1A

2

2

21

2

xx

x

2221

x

C

x

B

x

A

12122 22 xCxxBxAx

To find B we put A and C in and choose x to be any other number. Let x = 0

122212222)2( 22 CBA

C36 2C

13 -2

1022010203

120 22 B

223

42 B

3

42 B 3

2

6

4B

2 3

Let x = -2

Partial Fraction Decomposition With Quadratic Factors

When the denominator has a quadratic factor (that won’t factor), you need a fraction with a linear numerator.

41

12 xx 41 2

x

CBx

x

A(x+1)(x2 +4)

(x+1)(x2 +4)

(x+1)(x2 +4)

141 2 xCBxxA

The convenient x method doesn’t work as nicely on these kind so we’ll use the “equating coefficients” method. First multiply everything out.

41

12 xx 41 2

x

CBx

x

A

141 2 xCBxxA

Look at the kinds of terms on each side and equate coefficients (meaning put the coefficients = to each other)

CCxBxBxAAx 22 41Look at x2 terms: 0 = A + B

No x terms on left

Look at x terms: 0 = B + CLook at terms with no x’s: 1 = 4A + C

Solve these. Substitution would probably be easiest.

A = - B

C = - B

1 = 4(-B) + (-B)

5

1B

5

1A

5

1C

5

1

5

15

1

No x2 terms on left

Partial Fraction Decomposition With Repeated Quadratic Factors

When the denominator has a repeated quadratic factor (that won’t factor), you need a fraction with a linear numerator for each power.

22

23

4

x

xx

22244

x

DCx

x

BAx(x2 +4)2

(x2 +4)2

(x2 +4)2

DCxxBAxxx 4223multiply out

DCxBBxAxAxxx 44 2323

equate coefficients of various kinds of terms (next screen)

DCxBBxAxAxxx 44 2323

22

23

4

x

xx

22244

x

DCx

x

BAx

Look at x3 terms: 1 = A

Look at x2 terms: 1 = B

Look at x terms: 0 = 4A+C

Look at terms with no x:

0 = 4B+D

0 = 4(1)+C -4 =C

0 = 4(1)+D -4 = D

1 1 -4 -4

1 1

Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au