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Warm UpGraph

1

8)(

2

3

x

xxg

noneAV ..

xyAS ..

)0,2)(8,0(

2

2

2

2

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Partial Fraction

DecompositionChapter 2.7

Day 1 of 3

Tuesday, September 17, 2013

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Partial Fraction DecompositionThe purpose of PFD is to separate a fraction into

a sum of pre LCD terms.

)2)(4(

165

xx

x

2

3

4

2

xx

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Partial Fraction Decomposition1) Check if the fraction is proper (The power

in the numerator is less than that of the

denominator.) If it is not then divide. 6

72

xx

x

2) Factor the denominator.

3) Set up the factors. Dont

forget any special rules.

(Details to come.)

)2)(3(

7

xx

x

23)2)(3(7

xB

xA

xxx

4) Solve for A, B, etc. Use matching or

substitution method.

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Substitution Method4.1A) Multiply both sides by the LCD

2)2)(3(

3)2)(3(

)2)(3(

7)2)(3(

x

Bxx

x

Axx

xx

xxx

BxAxx )3()2(7

4.2A) Substitute a value for x that will eliminate A.

BA )32()22(72 2x

BA 505

B55

B1

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Substitution Method4.3A) Substitute a value for x that will eliminate B.

BA )33()23(73 3x

BA 0510

A2

5) Substitute A, B, etc. back into the original expansion.

23

x

B

x

A

2

1

3

2

xx

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Improper Fractions

xxxxxxx

23

234

262062

1) Check if the fraction is proper (The power in the numerator is

less than that of the denominator.) If it is not then divide.

62062 234 xxxxxxx 23 2

x

234 2 xxx ( )

6205 2 xx

xxx

xxx

23

2

2

62051A) Perform the PFD on theremainder over the divisor.

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Repeated Factor Rule2) Factor the denominator.

xxx

xx

23

2

2

6205

)12(

62052

2

xxx

xx2

2

)1(

6205

xx

xx

3) Set up the factors. Dont forget any special rules.

3A) The repeated factor rule. If a factor occurs more than

once then a term for each occurrence starting at power1 and up to power n must be placed.

22

2

)1(1)1(

6205

x

C

x

B

x

A

xx

xx

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Partial Fraction Decomposition4) Solve for A, B and C.

22

2

)1(1)1(

6205

x

C

x

B

x

A

xx

xx

4.1A) Multiply both sides by the LCD

2

222

2

22

)1()1(

1)1()1(

)1(

6205)1(

x

Cxx

x

Bxx

x

Axx

xx

xxxx

CBA )1()11)(1()11(6)1(20)1(5 22

4.2A) Substitute a value for x that will eliminate A.

xCBxxAxxx )1()1(6205 22

1x

CBA 006205

C9 C9

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Partial Fraction Decomposition4.3A) Substitute a value for x that will eliminate B.

CBA 0)10(0)10(6)0(20)0(5 22 0x

A6

4.4A) Substitute back A and C

xCBxxAxxx )1()1(6205 22

9)1(6)1(6205 22 xBxxxxx

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Partial Fraction Decomposition

4.5A) Substitute in any value for x but 0 and -1.

1x 9)1()11(16)11(6)1(20)1(522

B

92246205 B922431 B

B23331

B22

B1

5) Substitute A, B and C back into the original expansion.

2)1(

9

1

16

xxxx

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Partial Fraction Decomposition

2)2(

43

x

x2)2(2

x

B

x

A

BAxx )2(43

BA )22(4)2(32x

BA )0(46

B2

2)2(43 Axx

0x 2)20(4)0(3 A

224 A

A26

A3

2)2(

2

2

3

xx

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Check Your Work

2)2(

2

2

3

xx

22 )2(

2

)2(

)2(3

xx

x

22 )2(

2

)2(

63

xx

x

2)2(

263

x

x

2)2(

43

x

x