1. College Algebra Fifth Edition James Stewart Lothar Redlin Saleem Watson

2. Systems of Equations and Inequalities 6

3. Partial Fractions 6.4

4. Introduction To write a sum or difference of fractional expressions as a single fraction, we bring them to a common denominator.

5. Introduction However, for some applications of algebra to calculus, we must reverse this process. We must express a fraction such as 3x/(2x 2 – x – 1) as the sum of the simpler fractions 1/(x – 1) and 1/(2x + 1)

6. Partial Fractions These simpler fractions are called partial fractions. In this section, we learn how to find them.

7. Partial Fractions Let r be the rational function where the degree of P is less than the degree of Q.

8. Partial Fractions By the Linear and Quadratic Factors Theorem in Section 4.5, every polynomial with real coefficients can be factored completely into linear and irreducible quadratic factors. That is, factors of the form ax + b and ax 2 + bx + c where a, b, and c are real numbers.

9. Partial Fractions For instance, x 4 – 1 = (x 2 – 1)(x 2 + 1) = (x – 1)(x + 1)(x 2 + 1)

10. Partial Fraction Decomposition After we have completely factored the denominator Q of r, we can express r(x) as a sum of partial fractions of the form This sum is called the partial fraction decomposition of r.

11. Partial Fraction Decomposition Let’s examine the details of the four possible cases.

12. Case 1 The denominator is: A product of distinct linear factors.

13. Case 1 Suppose that we can factor Q(x) as Q(x) = (a 1 x + b 1 )(a 2 x + b 2 ) ··· (a n x + b n ) with no factor repeated. The partial fraction decomposition of P(x)/Q(x) takes the form

14. Partial Fraction Decomposition The constants A 1, A 2,..., A n are determined as in the following example.

15. E.g. 1—Distinct Linear Factors Find the partial fraction decomposition of:

16. E.g. 1—Distinct Linear Factors The denominator factors as: x 3 + 2x 2 – x – 2 = x 2 (x + 2) – (x + 2) = (x 2 – 1)(x + 2) = (x – 1)(x + 1)(x + 2) This gives the partial fraction decomposition

17. E.g. 1—Distinct Linear Factors Multiplying each side by the common denominator, (x – 1)(x + 1)(x + 2), we get: 5x + 7 = A(x + 1)(x + 2) + B(x – 1)(x + 2) + C(x – 1)(x + 1) = A(x 2 + 3x + 2) + B(x 2 + x – 2) + C(x 2 – 1) = (A + B + C)x 2 + (3A + B)x + (2A – 2B – C)

18. E.g. 1—Distinct Linear Factors If two polynomials are equal, their coefficients are equal. Thus, since 5x + 7 has no x 2 -term, we have A + B + C = 0. Similarly, by comparing the coefficients of x, we see that: 3A + B = 5 By comparing constant terms, we get: 2A – 2B – C = 7

19. E.g. 1—Distinct Linear Factors This leads to the following system of linear equations for A, B, and C. We use Gaussian elimination to solve this.

20. E.g. 1—Distinct Linear Factors

21. From the third equation, we get C = –1. Back-substituting, we find that B = –1 and A = 2 So, the partial fraction decomposition is:

22. Partial Fraction Decomposition The same approach works in the remaining cases. We set up the partial fraction decomposition with the unknown constants, A, B, C,.... Then, we multiply each side of the resulting equation by the common denominator, simplify the right-hand side of the equation, and equate coefficients.

23. Partial Fractions This gives a set of linear equations that will always have a unique solution. This is provided that the partial fraction decomposition has been set up correctly.

24. Case 2 The denominator is: A product of linear factors, some of which are repeated.

25. Case 2 Suppose the complete factorization of Q(x) contains the linear factor ax + b repeated k times—that is, (ax + b) k is a factor of Q(x). Then, corresponding to each such factor, the partial fraction decomposition for P(x)/Q(x) contains

26. E.g. 2—Repeated Linear Factors Find the partial fraction decomposition of: The factor x – 1 is repeated three times in the denominator.

27. E.g. 2—Repeated Linear Factors So, the partial fraction decomposition has the form

28. E.g. 2—Repeated Linear Factors We then multiply each side by the common denominator x(x – 1) 3. x 2 + 1 = A(x – 1) 3 + Bx(x – 1) 2 + Cx(x – 1) + Dx = A(x 3 – 3x 2 + 3x – 1) + B(x 3 – 2x 2 + x) + C(x 2 – x) + Dx = (A + B)x 3 + (–3A – 2B +C)x 2 + (3A + B – C + D)x – A

29. E.g. 2—Repeated Linear Factors Equating coefficients, we get: If we rearrange these by putting the last one in the first position, we can easily see (using substitution) that the solution to the system is: A = –1, B = 1, C = 0, D = 2

30. E.g. 2—Repeated Linear Factors So, the partial fraction decomposition is:

31. Case 3 The denominator has: Irreducible quadratic factors, none of which is repeated.

32. Case 3 Suppose the complete factorization of Q(x) contains the quadratic factor ax 2 + bx + c (which can’t be factored further). Then, corresponding to this, the partial fraction decomposition of P(x)/Q(x) will have a term of the form

33. E.g. 3—Distinct Quadratic Factors Find the partial fraction decomposition of: Since x 3 + 4x = x(x 2 + 4), which can’t be factored further, we write:

34. E.g. 3—Distinct Quadratic Factors Multiplying by x(x 2 + 4), we get: 2x 2 – x + 4 = A(x 2 + 4) + (Bx + C)x = (A + B)x 2 + Cx + 4A

35. E.g. 3—Distinct Quadratic Factors Equating coefficients gives us: So, A = 1, B = 1, and C = –1.

36. E.g. 3—Distinct Quadratic Factors The required partial fraction decomposition is:

37. Case 4 The denominator has: A repeated irreducible quadratic factor.

38. Case 4 Suppose the complete factorization of Q(x) contains the factor (ax 2 + bx + c) k, where ax 2 + bx + c can’t be factored further. Then, the partial fraction decomposition of P(x)/Q(x) will have the terms

39. E.g. 4—Repeated Quadratic Factors Write the form of the partial fraction decomposition of:

40. E.g. 4—Repeated Quadratic Factors

41. Repeated Quadratic Factors To find the values of A, B, C, D, E, F, G, H, I, J, and K in Example 4, we would have to solve a system of 11 linear equations. Although possible, this would certainly involve a great deal of work!

42. Using Long Division The techniques we have described in this section apply only to: Rational functions P(x)/Q(x) in which the degree of P is less than the degree of Q. If this isn’t the case, we must first use long division to divide Q into P.

43. E.g. 5—Long Division to Prepare for Partial Fractions Find the partial fraction decomposition of: The degree of the numerator is larger than that of the denominator.

44. E.g. 5—Long Division to Prepare for Partial Fractions So, we use long division.

45. E.g. 5—Long Division to Prepare for Partial Fractions Thus, we obtain: The remainder term now satisfies the requirement that the degree of the numerator is less than that of the denominator.

46. E.g. 5—Long Division to Prepare for Partial Fractions At this point, we proceed as in Example 1 to obtain the decomposition