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: 38

Answer

$= \frac{(3k+4)(k-3)}{2(k-4)(k+5)}$

Work Step by Step

$\frac{3k^{2}+19k+20}{k^{2}+6k+5} \div \frac{2k^{2}+2k-40}{k^{2}-2k-3}$ $= \frac{3k^{2}+19k+20}{k^{2}+6k+5} \times \frac{k^{2}-2k-3}{2k^{2}+2k-40}$ $= \frac{3k^{2}+19k+20}{k^{2}+6k+5} \times \frac{k^{2}-2k-3}{2(k^{2}+k-20)}$ $= \frac{3k^{2}+15k+4k+20}{(k+5)(k+1)} \times \frac{(k-3)(k+1)}{2(k-4)(k+5)}$ $= \frac{3k(k+5)+4(k+5)}{(k+5)(k+1)} \times \frac{(k-3)(k+1)}{2(k-4)(k+5)}$ $= \frac{(3k+4)(k+5)}{(k+5)(k+1)} \times \frac{(k-3)(k+1)}{2(k-4)(k+5)}$ $= \frac{(3k+4)}{(k+1)} \times \frac{(k-3)(k+1)}{2(k-4)(k+5)}$ $= \frac{(3k+4)}{1} \times \frac{(k-3)}{2(k-4)(k+5)}$ $= \frac{(3k+4)(k-3)}{2(k-4)(k+5)}$
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