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: 37

Answer

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

Work Step by Step

$\frac{c^{2}-c-20}{c^{2}+6c+8} \div \frac{2c^{2}+11c+12}{3c^{2}+c-10}$ $= \frac{c^{2}-c-20}{c^{2}+6c+8} \times \frac{3c^{2}+c-10}{2c^{2}+11c+12}$ $= \frac{(c-5)(c+4)}{(c+4)(c+2)} \times \frac{3c^{2}-5c+6c-10}{2c^{2}+8c+3c+12}$ $= \frac{(c-5)}{(c+2)} \times \frac{c(3c-5)+2(3c-5)}{2c(c+4)+3(c+4)}$ $= \frac{(c-5)}{(c+2)} \times \frac{(c+2)(3c-5)}{(2c+3)(c+4)}$ $= \frac{(c-5)}{1} \times \frac{(3c-5)}{(2c+3)(c+4)}$ $= \frac{(3c-5)(c-5)}{(2c+3)(c+4)}$
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