1. Copyright © Cengage Learning. All rights reserved. 7 Systems of Equations and Inequalities

2. Copyright © Cengage Learning. All rights reserved. Partial Fractions 7.4

3. 3  Recognize partial fraction decompositions of rational expressions.  Find partial fraction decompositions of rational expressions. Objectives

4. 4 Introduction

5. 5 In this section, you will learn to write a rational expression as the sum of two or more simpler rational expressions. For example, the rational expression can be written as the sum of two fractions with first-degree denominators. That is,

6. 6 Introduction Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition of the left side.

7. 7 Introduction

8. 8 Partial Fraction Decomposition

9. 9 The following example demonstrate algebraic techniques for determining the constants in the numerators of partial fractions. Note that the techniques vary slightly, depending on the type of factors of the denominator: linear or quadratic, distinct or repeated.

10. 10 Example 1 – Distinct Linear Factors Write the partial fraction decomposition of. Solution: The expression is proper, so you should begin by factoring the denominator. Because x 2 – x – 6 = (x – 3)(x + 2), you should include one partial fraction with a constant numerator for each linear factor of the denominator. Write the form of the decomposition as follows. Write form of decomposition.

11. 11 Example 1 – Solution Multiplying each side of this equation by the least common denominator, (x – 3)(x + 2), leads to the basic equation x + 7 = A(x + 2) + B(x – 3). Because this equation is true for all x, substitute any convenient values of x that will help determine the constants A and B. Values of x that are especially convenient are those that make the factors (x + 2) and (x – 3) equal to zero. cont’d Basic equation

12. 12 Example 1 – Solution For instance, to solve for B, let x = –2. Then, –2 + 7 = A(–2 + 2) + B(–2 – 3) 5 = A(0) + B(–5) 5 = – 5B –1 = B. To solve for A, let x = 3 and obtain 3 + 7 = A(3 + 2) + B(3 – 3) 10 = A(5) + B(0) Substitute –2 for x. Substitute 3 for x. cont’d

13. 13 Example 1 – Solution 10 = 5A 2 = A. So, the partial fraction decomposition is Check this result by combining the two partial fractions on the right side of the equation, or by using your graphing utility. cont’d

14. 14 Partial Fraction Decomposition The procedure used to solve for the constants in Example1 works well when the factors of the denominator are linear. However, when the denominator contains irreducible quadratic factors, you should use a different procedure, which involves writing the right side of the basic equation in polynomial form, equating the coefficients of like terms, and using a system of equations to solve for the coefficients.

15. 15 Partial Fraction Decomposition

16. 16 Partial Fraction Decomposition Keep in mind that for improper rational expressions you must first divide before applying partial fraction decomposition.