Intermediate Algebra (6th Edition)

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

Chapter 6 - Section 6.1 - Rational Functions and Multiplying and Dividing Rational Expressions - Exercise Set - Page 346: 67

Answer

$\dfrac{5(3a+2)}{a}$

Work Step by Step

Factoring the expressions and cancelling the common factors between the numerator and the denominator, the given expression, $ \dfrac{5a^2-20}{3a^2-12a}\div\dfrac{a^3+2a^2}{2a^2-8a}\cdot\dfrac{9a^3+6a^2}{2a^2-4a} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{5a^2-20}{3a^2-12a}\cdot\dfrac{2a^2-8a}{a^3+2a^2}\cdot\dfrac{9a^3+6a^2}{2a^2-4a} \\\\ \dfrac{5(a^2-4)}{3a(a-4)}\cdot\dfrac{2a(a-4)}{a^2(a+2)}\cdot\dfrac{3a^2(3a+2)}{2a(a-2)} \\\\ \dfrac{5(a+2)(a-2)}{3a(a-4)}\cdot\dfrac{2a(a-4)}{a^2(a+2)}\cdot\dfrac{3a^2(3a+2)}{2a(a-2)} \\\\ \dfrac{5(\cancel{a+2})(\cancel{a-2})}{\cancel{3}a(\cancel{a-4})}\cdot\dfrac{\cancel{2a}(\cancel{a-4})}{\cancel{a^2}(\cancel{a+2})}\cdot\dfrac{\cancel{3}\cancel{a^2}(3a+2)}{\cancel{2a}(\cancel{a-2})} \\\\= \dfrac{5(3a+2)}{a} .\end{array}
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