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

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

3. OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Partial-Fraction Decomposition Learn to decompose where Q(x) has distinct irreducible quadratic factors. Learn to decompose where Q(x) has repeated irreducible quadratic factors. SECTION 8.3 4 5 Continued

4. Slide 8.3- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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.

5. Slide 8.3- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley CASE 1: THE DENOMINATOR IS THE PRODUCT OF DISTINCT (NONREPEATED) LINEAR FACTORS Suppose Q(x) can be factored as where A 1, A 2, …, A n, are constants to be determined. with no factor repeated. The partial-fraction decomposition of is of the form

6. Slide 8.3- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR PARTIAL-FRACTION DECOMPOSITION Step 1Write the form of the partial-fraction decomposition with the unknown constants A, B, C, … in the numerators of the decomposition. Step 2Multiply both sides of the equation in Step 1 by the original denominator. Use the distributive property and eliminate common factors. Simplify.

7. Slide 8.3- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR PARTIAL-FRACTION DECOMPOSITION Step 3Write both sides of the equation in Step 2 in descending powers of x, and equate the coefficients of like powers of x. Step 4Step 3 will result in a system of linear equations in A, B, C,…. Solve this linear system for the constants A, B, C,…. Step 5Substitute the values for A, B, C,… obtained in Step 4 into the equation in Step 1, and write the partial-fraction decomposition.

8. Slide 8.3- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors Find the partial-fraction decomposition of the expression. Solution Factor the denominator. Step 1Write the partial-fraction.

9. Slide 8.3- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors Solution continued Step 2Multiply by original denominator. Distribute

10. Slide 8.3- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors Solution continued Step 3Now use the fact that two polynomials are equal if and only if the coefficients of the like powers are equal.

11. Slide 8.3- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors Solution continued Step 4Solve the system of equations in Step 3 to obtain A = 1, B = 2, and C = –3. Equating corresponding coefficients leads to the system of equations.

12. Slide 8.3- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors Solution continued Step 5Since 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.

13. Slide 8.3- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors 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

14. Slide 8.3- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors 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

15. Slide 8.3- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Partial-Fraction Decomposition When the Denominator has Only Distinct Linear Factors Alternative Solution continued Thus, A = 1, B = 2 and C = –3 and the partial- fraction decomposition is given by

16. Slide 8.3- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley CASE 2: THE DENOMINATOR HAS A REPEATED LINEAR FACTOR where A 1, A 2, …, A n, are constants to be determined. 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 is

17. Slide 8.3- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors 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

18. Slide 8.3- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors Solution continued Step 2Multiply by original denominator. Distribute

19. Slide 8.3- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors Solution continued Substitute x = 1 (because it causes the terms with A and B to be 0) in the equation to obtain Substitute x = –3 (because it causes the terms with B and C to be 0) to get

20. Slide 8.3- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors Solution continued To obtain the value of B, we replace x with any convenient number, say, 0. We have Now substituteand

21. Slide 8.3- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors Solution continued Substituteand in the decomposition in Step 1 to obtain the partial-fraction decomposition

22. Slide 8.3- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Finding the Partial-Fraction Decomposition When the Denominator has Repeated Linear Factors Solution continued or

23. Slide 8.3- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley CASE 3: THE DENOMINATOR HAS A DISTINCT(NONREPEATED) IRREDUCIBLE QUADRATIC FACTOR where A and B are constants to be determined. Suppose ax 2 + b + c is an irreducible quadratic factor of Q(x). Then the portion of that corresponds to ax 2 + bx + c has the form the partial-fraction decomposition of

24. Slide 8.3- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor Find the partial-fraction decomposition of Solution Step 1(x – 4) is linear, (x 2 + 1) is irreducible; thus, the partial-fraction decomposition has the form

25. Slide 8.3- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor Solution continued Step 2Multiply both sides of the decomposition in Step 1 by the original denominator, and simplify to obtain Substitute x = 4 to obtain 17 = 17A, or A = 1. Step 3Collect like terms, write both sides of the equation in descending powers of x.

26. Slide 8.3- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor Solution continued Equate coefficients of the like powers of x to obtain Step 4Substitute 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.

27. Slide 8.3- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor Solution continued Step 5Substitute A = 1, B = 2, and C = 0 into the decomposition in Step 1 to obtain

28. Slide 8.3- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley CASE 4: THE DENOMINATOR HAS A REPEATED IRREDUCIBLE QUADRATIC FACTOR Suppose the denominator Q(x) has a factor (ax 2 + bx + c) m where m ≥ 2 is an integer and ax 2 + bx + c is irreducible. Then the portion that corresponds to the factor ax 2 + bx + c has the form of the partial-fraction decomposition of

29. Slide 8.3- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley CASE 4: THE DENOMINATOR HAS A REPEATED IRREDUCIBLE QUADRATIC FACTOR where A 1, B 1, A 2, B 2, …, A m, B m are constants to be determined.

30. Slide 8.3- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Irreducible Quadratic Factor Find the partial-fraction decomposition of Solution Step 1(x – 1) is a nonrepeating linear factor, (x 2 + 4) is an irreducible quadratic factor repeated twice, the partial- fraction decomposition has the form

31. Slide 8.3- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Irreducible Quadratic Factor Solution continued Step 2Multiply both sides by the original denominator.

32. Slide 8.3- 32 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Irreducible Quadratic Factor Solution continued Distribute and eliminate common factors Distribute

33. Slide 8.3- 33 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Irreducible Quadratic Factor Solution continued Substitute x = 1 and simplify to obtain 25 = 25A, or A = 1. Step 3Multiply the right side of equation in Step 2 and collect like terms to obtain Equate coefficients of the like powers of x to obtain

34. Slide 8.3- 34 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Irreducible Quadratic Factor Solution continued Back substitute A = 1 in Equation (1) to obtain B = 1. Back substitute B = 1 into Equation (2) to get C = 0.

35. Slide 8.3- 35 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Irreducible Quadratic Factor Solution continued Step 5Substitute A = 1, B = 1, C = 0, D = 1, E = 3 Back substitute A = 1 and C = 0 in Equation (5) to obtain E = 3. Back substitute A = 1, B = 1 and C = 0 into Equation (3) to get D = 1.