Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.3 Multiplying and Dividing Rational Expressions - 7.3 Exercises - Page 581: 36

Answer

$ = \frac{(k+5)(k-7)}{(k+7)(k+3)}$

Work Step by Step

$\frac{k^{2}+k-20}{k^{2}+10k+21} \div \frac{k^{2}-10k+24}{k^{2}-13k+42}$ $ = \frac{k^{2}+k-20}{k^{2}+10k+21} \times \frac{k^{2}-13k+42}{k^{2}-10k+24}$ $ = \frac{(k-4)(k+5)}{(k+7)(k+3)} \times \frac{(k-6)(k-7)}{(k-6)(k-4)}$ $ = \frac{(k-4)(k+5)}{(k+7)(k+3)} \times \frac{(k-7)}{(k-4)}$ $ = \frac{(k+5)}{(k+7)(k+3)} \times \frac{(k-7)}{1}$ $ = \frac{(k+5)(k-7)}{(k+7)(k+3)}$
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