1 / 46

Introduction to Partial Fractions: Simplifying Fractional Expressions

Learn how to simplify fractional expressions by writing them as a sum or difference of simpler fractions, known as partial fractions. This introduction covers cases with distinct linear factors, repeated linear factors, and irreducible quadratic factors.

sperkins
Download Presentation

Introduction to Partial Fractions: Simplifying Fractional Expressions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related