Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Mid-Chapter Check Point - Page 853: 14



Work Step by Step

Here, we have $\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}=\dfrac{A}{x+2}+\dfrac{B}{x+1}+\dfrac{C}{x-1}$ $10x^2+9x-7 =A(x-1)(x+1)+B(x+2)(x-1)+C(x+2)(x+1)$ or, $10x^2+9x-7 =(A+B+C)x^2+(B+3C)x-A-2B+2C$ Equating the coefficients, we get $A+B+C=10$ and $B+3C =9 \implies B=-2$ and $-A-2B+2C=-7$ $ \implies (A+B+C)+(-A-2B+2C)=10-7$ gives: $C=2$ and $ B+3(2)=9 \implies B=3$ $A+3+2= 10 \implies A=5$ Hence, $\dfrac{10x^2+9x-7}{(x+2)(x^2-1)}=\dfrac{5}{x+2}+\dfrac{3}{x+1}+\dfrac{2}{x-1}$
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