Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 79

Answer

$\dfrac{3(x+2)^{2}(x-3)^{2}-(x+2)^{3}(2)(x-3)}{(x-3)^{4}}=\dfrac{(x+2)^{2}(x-13)}{(x-3)^{3}}$

Work Step by Step

$\dfrac{3(x+2)^{2}(x-3)^{2}-(x+2)^{3}(2)(x-3)}{(x-3)^{4}}$ Take out common factor $(x+2)^{2}(x-3)$ from the numerator: $\dfrac{3(x+2)^{2}(x-3)^{2}-(x+2)^{3}(2)(x-3)}{(x-3)^{4}}=...$ $...=\dfrac{(x+2)^{2}(x-3)[3(x-3)-2(x+2)]}{(x-3)^{4}}=...$ Simplify the expression inside brackets in the numerator: $...=\dfrac{(x+2)^{2}(x-3)[3x-9-2x-4]}{(x-3)^{4}}=...$ $...=\dfrac{(x+2)^{2}(x-3)(x-13)}{(x-3)^{4}}=...$ Simplify the rational expression: $...=\dfrac{(x+2)^{2}(x-13)}{(x-3)^{3}}$
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