SECTION 11.5 Partial Fraction Decomposition 819 Solution The denominator contains the repeated irreducible quadratic factor ( ) + x 4 , 2 2 so by Case 4, ( ) ( ) + + = + + + + + x x x Ax B x Cx D x 4 4 4 3 2 2 2 2 2 2 (11) Clear fractions to obtain ( ) ( ) + = + + + + x x Ax B x Cx D 4 3 2 2 Expanding and collecting like terms yields the identity ( ) + = + + + + + x x Ax Bx A C x B D 4 4 3 2 3 2 Equating coefficients results in the system = = + = + = ⎧ ⎨ ⎪⎪ ⎪⎪ ⎪ ⎩ ⎪⎪ ⎪⎪ ⎪ A B A C B D 1 1 4 0 4 0 The solution is = = = − = − A B C D 1, 1, 4, 4. From equation (11), ( ) ( ) + + = + + + − − + x x x x x x x 4 1 4 4 4 4 3 2 2 2 2 2 2 Now Work PROBLEM 39 ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 11.5 Assess Your Understanding 1. True or False The equation ( ) ( ) − − = − x x x 1 1 2 2 is an example of an identity. (p. A44) 2. True or False The rational expression − + x x 5 1 1 2 3 is proper. (p. 241 ) 3. Reduce to lowest terms: − − x x 3 12 16 2 (pp. A35–A36) 4. True or False Every polynomial with real numbers as coefficients can be factored into a product of linear and/or irreducible quadratic factors. (p. 233) Skill Building In Problems 5–16, determine whether the given rational expression is proper or improper. If the expression is improper, rewrite it as the sum of a polynomial and a proper rational expression. 5. − x x 1 2 6. + − x x 5 2 1 3 7. + − x x 5 4 2 2 8. − − x x 3 2 1 2 2 9. + − + + − x x x x x 12 9 2 15 3 2 2 10. − − − − x x x x 6 5 7 3 2 5 3 2 11. − − + x x x x 5 7 6 2 3 12. + − − x x x x x 12 9 9 3 2 2 4 13. + − − x x x 5 2 1 4 3 2 14. + − + x x x 3 2 8 4 2 3 15. ( ) ( )( ) − + − x x x x 1 4 3 16. ( ) + + x x x 2 4 1 2 2 1. Now Work 1. Modeling 1.ExplainingConcepts Calculus Preview 1.InteractiveFigure
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