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: 13



Work Step by Step

Here, we have $\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}+\dfrac{C}{(x-2)^3}$ $x^2-6x+3 =A(x-2)^2+B(x-2)+C$ or, $x^2-6x+3 =Ax^2+(-4A+B)x+4A-2B+C$ Equating the coefficients, we get $A=1$ and $ -4A+B = -6 \implies B=-2$ and $4A-2B+C =3 \implies C=-5$ Hence, $\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{1}{x-2}-\dfrac{2}{(x-2)^2}-\dfrac{5}{(x-2)^3}$
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