Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Test: 16

Answer

$\dfrac{(x-3)^2}{x-2}$

Work Step by Step

The given expression, $ \dfrac{\dfrac{x^2-5x+6}{x+3}}{\dfrac{x^2-4x+4}{x^2-9}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{x^2-5x+6}{x+3}\div\dfrac{x^2-4x+4}{x^2-9} \\\\= \dfrac{x^2-5x+6}{x+3}\cdot\dfrac{x^2-9}{x^2-4x+4} \\\\= \dfrac{(x-2)(x-3)}{x+3}\cdot\dfrac{(x+3)(x-3)}{(x-2)(x-2)} \\\\= \dfrac{(\cancel{x-2})(x-3)}{\cancel{x+3}}\cdot\dfrac{(\cancel{x+3})(x-3)}{(\cancel{x-2})(x-2)} \\\\= \dfrac{(x-3)^2}{x-2} .\end{array}
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