Slide 8.3- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Partial-Fraction Decomposition

Learn the definition of partial fraction

decomposition.

Learn to decompose where Q(x) has only

distinct linear factors.

Learn to decompose where Q(x) has

repeated linear factors.

SECTION 8.3

1

2

3

P x Q x

P x Q x

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Partial-Fraction Decomposition


distinct irreducible quadratic factors.


repeated irreducible quadratic factors.

SECTION 8.3

4

5

P x Q x

P x Q x

Continued


PARTIAL FRACTIONS

Each of the two fractions on the right is called a partial fraction. Rewriting a single fraction as the sum of two (or more) fractions is called the partial-fraction decomposition of the rational expression.

5x 7

x 1 x 3 2

x 3

3

x 1


CASE 1: THE DENOMINATOR IS THE PRODUCT OF DISTINCT (NONREPEATED)

LINEAR FACTORSSuppose Q(x) can be factored as

where A1, A2, …, An, are constants to be determined.

Q x x a1 x a2 ... x an ,

P x Q x

A1

x a1

A2

x a2

...An

x an

,

with no factor repeated. The partial-fraction

decomposition of is of the formP x Q x


PROCEDURE FOR PARTIAL-FRACTION DECOMPOSITION

Step 1 Write the form of the partial-fraction decomposition with the unknown constants A, B, C, … in the numerators of the decomposition.

Step 2 Multiply both sides of the equation in Step 1 by the original denominator. Use the distributive property and eliminate common factors. Simplify.


PROCEDURE FOR PARTIAL-FRACTION DECOMPOSITION

Step 3 Write both sides of the equation in Step 2 in descending powers of x, and equate the coefficients of like powers of x.

Step 4 Step 3 will result in a system of linear equations in A, B, C,… . Solve this linear system for the constants A, B, C,… .

Step 5 Substitute the values for A, B, C,… obtained in Step 4 into the equation in Step 1, and write the partial-fraction decomposition.


EXAMPLE 1Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors

10x 4

x3 4x

Find the partial-fraction decomposition of the expression.

x3 4x x x2 4 x x 2 x 2

Solution

Factor the denominator.

Step 1 Write the partial-fraction.

10x 4

x x 2 x 2 A

x

B

x 2

C

x 2



Solution continuedStep 2 Multiply by original denominator.

x x 2 x 2 10x 4

x x 2 x 2

x x 2 x 2 A

x

B

x 2

C

x 2

Distribute

10x 4 A x 2 x 2 Bx x 2 Cx x 2



Solution continued

Step 3 Now use the fact that two polynomials are equal if and only if the coefficients of the like powers are equal.

10x 4 A x 2 x 2 Bx x 2 Cx x 2 A x2 4 B x2 2x C x2 2x Ax2 4A Bx2 2Bx Cx2 2Cx

10x 4 A B C x2 2B 2C x 4A

0x2 10x 4 A B C x2 2B 2C x 4A



Solution continued

Step 4 Solve the system of equations in Step 3 to obtain A = 1, B = 2, and C = –3.

A B C 0

2B 2C 10

4A 4

Equating corresponding coefficients leads to the system of equations.



Solution continued

Step 5 Since A = 1, B = 2, and C = –3, the partial-fraction decomposition is

An alternative (and sometimes quicker) method of finding the constants is to substitute well-chosen values for x in the equation (identity) obtained in Step 2.

10x 4

x3 4x

1

x

2

x 2

3

x 2

1

x

2

x 2

3

x 2



Alternative Solution

Start with equation (1) from Step 2.

Substitute x = 2 (because it causes the terms with A and C to be 0) in equation (1) to obtain

10x 4 A x 2 x 2 Bx x 2 Cx x 2

10 2 4

A 2 2 2 2 B 2 2 2 C 2 2 2 16 8B

2 B



Alternative Solution continued

Substitute x = –2 (because it causes the terms with A and B to be 0) in equation (1) to obtain

Substitute x = 0 (because it causes the terms with B and C to be 0) in equation (1) to obtain

10 2 4

A 2 2 2 2 B 2 2 2 C 2 2 2 24 8C

3 C



Alternative Solution continued

Thus, A = 1, B = 2 and C = –3 and the partial-fraction decomposition is given by

4 4A

1 A

10 0 4

A 0 2 0 2 B 0 0 2 C 0 0 2

10x 4

x3 4x

1

x

2

x 2

3

x 2

1

x

2

x 2

3

x 2


CASE 2: THE DENOMINATOR HAS A REPEATED LINEAR FACTOR

where A1, A2, …, An, are constants to be determined.

A1

x a

A2

x a 2 ...An

x a m ,

Let (x – a)m be the linear factor (x – a) that is repeated m times in Q(x). Then the portion of the partial-fraction decomposition of

that corresponds to the factor (x – a)m isP x Q x


EXAMPLE 2Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors

x 4

x 3 x 1 2

Find the partial-fraction decomposition of the expression.

Solution

Step 1 (x – 1) is repeated twice, (x + 3) is nonrepeating, the partial-fraction decomposition has the form

x 4

x 3 x 1 2 A

x 3

B

x 1

C

x 1 2



Solution continuedStep 2 Multiply by original denominator.

x 3 x 1 2 x 4

x 3 x 1 2

x 3 x 1 2 A

x 3

B

x 1

C

x 1 2

Distribute

x 4 A x 1 2 B x 3 x 1 C x 3



Solution continuedSubstitute x = 1 (because it causes the terms with A and B to be 0) in the equation to obtain

5 4C

C 5

4

1 4 A 1 1 2 B 1 3 1 1 C 1 3

Substitute x = –3 (because it causes the terms with B and C to be 0) to get

3 4 A 3 1 2 B 3 3 3 1 C 3 3



Solution continued1 16A

A 1

16

To obtain the value of B, we replace x with any convenient number, say, 0. We have

0 4 A 0 1 2 B 0 3 0 1 C 0 3 4 A 3B 3C

A 1

16C

5

4.Now substitute and



Solution continued

4 1

16 3B 3

5

4

64 1 48B 60

B 1

16

Substitute A 1

16, C

5

4andB

1

16,

in the decomposition in Step 1 to obtain the partial-fraction decomposition



Solution continued

x 4

x 3 x 1 2

116

x 3

1

16x 1

54

x 1 2

or

x 4

x 3 x 1 2 1

16 x 3 1

16 x 1 5

4 x 1 2


CASE 3: THE DENOMINATOR HAS A DISTINCT(NONREPEATED) IRREDUCIBLE

QUADRATIC FACTOR

where A and B are constants to be determined.

Ax B

ax2 bx c,

Suppose ax2 + b + c is an irreducible quadratic factor of Q(x). Then the portion of

that corresponds to ax2 + bx + c has the form

P x Q x the partial-fraction decomposition of


EXAMPLE 3Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor

3x2 8x 1

x 4 x2 1 .Find the partial-fraction decomposition of

Solution

Step 1 (x – 4) is linear, (x2 + 1) is irreducible; thus, the partial-fraction decomposition has the form

3x2 8x 4

x 4 x2 1 A

x 4

Bx C

x2 1.



Solution continuedStep 2 Multiply both sides of the

decomposition in Step 1 by the original denominator, and simplify to obtain

3x2 8x 1 A x2 1 Bx C x 4

Substitute x = 4 to obtain 17 = 17A, or A = 1.

Step 3 Collect like terms, write both sides of the equation in descending powers of x.

3x2 8x 1 A B x2 4B C x A 4C



Solution continuedEquate coefficients of the like powers of x to obtain

Step 4 Substitute A = 1 in equation (1) to obtain 1 + B = 3, or B = 2.Substitute A = 1 in equation (3) to obtain 1 – 3C = 1, or C = 0.

A B 3 (1)

4B C 8 (2)

A 4C 1 (3)



Solution continued

Step 5 Substitute A = 1, B = 2, and C = 0 into the decomposition in Step 1 to obtain

3x2 8x 4

x 4 x2 1 1

x 4

2x

x2 1.


CASE 4: THE DENOMINATOR HAS A REPEATED IRREDUCIBLE QUADRATIC

FACTOR

Suppose the denominator Q(x) has a factor (ax2 + bx + c)

m where m ≥ 2 is an integer and ax2 + bx + c is irreducible. Then the portion

that corresponds to the factor ax2 + bx + c has the form

P x Q x of the partial-fraction decomposition of


CASE 4: THE DENOMINATOR HAS A REPEATED IRREDUCIBLE QUADRATIC

FACTOR

where A1, B1, A2, B2, …, Am, Bm are constants to be determined.

A1x B1

ax2 bx c

A2 x B2

ax2 bx c 2 ...Am x Bm

ax2 bx c m ,


EXAMPLE 4Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Irreducible Quadratic Factor

2x4 x3 13x2 2x 13

x 1 x2 4 2 .

Find the partial-fraction decomposition of

Solution

Step 1 (x – 1) is a nonrepeating linear factor, (x2 + 4) is an irreducible quadratic factor repeated twice, the partial-fraction decomposition has the form



Solution continued

Step 2 Multiply both sides by the original denominator.

2x4 x3 13x2 2x 13

x 1 x2 4 2

A

x 1 Bx C

x2 4

Dx E

x2 4 2



Solution continued

Distribute and eliminate common factors

2x4 x3 13x2 2x 13

x 1 x2 4 2 A

x 1 Bx C

x2 4

Dx E

x2 4 2

Distribute

2x4 x3 13x2 2x 13

A x2 4 2 Bx C x 1 x2 4 Dx E x 1



Solution continuedSubstitute x = 1 and simplify to obtain 25 = 25A, or A = 1.

Step 3 Multiply the right side of equation in Step 2 and collect like terms to obtain

2x4 x3 13x2 2x 13

A B x4 B C x3 8A 4B C D x2

4B 4C D E x 16A 4C E Equate coefficients of the like powers of x to obtain



Solution continued

A B 2 (1)

B C 1 (2)

8A 4B C D 13 (3)

4B 4C D E 2 (4)

16A 4C E 13 (5)

Back substitute A = 1 in Equation (1) to obtain B = 1. Back substitute B = 1 into Equation (2) to get C = 0.



Solution continued

Step 5 Substitute A = 1, B = 1, C = 0, D = 1, E = 3

Back substitute A = 1 and C = 0 in Equation (5) to obtain E = 3.

2x4 x3 13x2 2x 13

x 1 x2 4 2 1

x 1 x

x2 4

x 3

x2 4 2 .

Back substitute A = 1, B = 1 and C = 0 into Equation (3) to get D = 1.

Slide 8.3- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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Transcript of Slide 8.3- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.