PAR TIALFRAC TION+

DECOMPOSITION

23

34

xx

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

22

xx

33

xx

329384

xxxx

617

2 xxx 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

617

2 xxx We start by factoring the denominator.

2317

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

xB

xA

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

617

2 xxx

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

2317

xx

x23

xB

xA

4A

4 3

617

2 xxx

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

212

xx

x

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

2221

xC

xB

xA(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 31

A

2

2

212

xx

x

2221

xC

xB

xA

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

1022010203120 22 B

22342 B

342 B 3

264

B

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.

411

2 xx 41 2

x

CBxxA(x+1)(x

2 +4)(x+1)(x

2 +4)(x+1)(x

2 +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.

411

2 xx 41 2

x

CBxxA

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)51

B

51

A51

C

51

51

51

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

xxx

222 44

x

DCxx

BAx(x2 +4)2 (x2 +4)2(x2 +4)2

DCxxBAxxx 4223 multiply out

DCxBBxAxAxxx 44 2323

equate coefficients of various kinds of terms (next screen)

DCxBBxAxAxxx 44 2323

22

23

4

xxx

222 44

x

DCxx

BAx

Look at x3 terms: 1 = A

Look at x2 terms: 1 = B

Look at x terms: 0 = 4A+CLook 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