Introduction to Partial Fractions: Simplifying Fractional Expressions

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/(2x2 – 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 functionwhere 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 ax2 + bx +c where a, b, and c are real numbers.

9. Partial Fractions • For instance, x4 – 1 = (x2 – 1)(x2 + 1) = (x – 1)(x + 1)(x2 + 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 fractionsof the form • This sum is called the partial fraction decompositionof 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) asQ(x) = (a1x + b1 )(a2x + b2 ) ··· (anx + bn)with no factor repeated. • The partial fraction decomposition of P(x)/Q(x) takes the form

14. Partial Fraction Decomposition • The constants A1, A2, . . . , Anare 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:x3 + 2x2 – x – 2 = x2(x + 2) – (x + 2) = (x2 – 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(x2 + 3x + 2) + B(x2 + x – 2) + C(x2 – 1) = (A + B + C)x2 + (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 x2-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. E.g. 1—Distinct Linear Factors • 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. • x2 + 1 = A(x – 1)3 + Bx(x – 1)2 + Cx(x – 1) + Dx =A(x3 – 3x2 + 3x – 1) + B(x3 – 2x2 + x) + C(x2 – x) + Dx = (A + B)x3 + (–3A – 2B +C)x2 + (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 ax2 + 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 x3 + 4x =x(x2 + 4), which can’t be factored further, we write:

34. E.g. 3—Distinct Quadratic Factors • Multiplying by x(x2 + 4), we get: 2x2 – x + 4 = A(x2 + 4) + (Bx + C)x = (A +B)x2 + 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 (ax2 + bx + c)k, where ax2 + 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