College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Review Exercises - Page 76: 77



Work Step by Step

Factoring the expressions and then cancelling the common factors between the numerator and the denominator, the given expression, $ \dfrac{k^2+k}{8k^3}\cdot\dfrac{4}{k^2-1} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{k(k+1)}{8k^3}\cdot\dfrac{4}{(k+1)(k-1)} \\\\= \dfrac{\cancel{k}(\cancel{k+1})}{\cancel{4}(2)\cancel{k}\cdot k^2}\cdot\dfrac{\cancel{4}}{(\cancel{k+1})(k-1)} \\\\= \dfrac{1}{2k^2(k-1)} .\end{array}
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