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 348: 108



Work Step by Step

Multiplying by the reciprocal of the divisor, the given expression, $ \dfrac{y^{2n}+7y^{n}+10}{10}\div\dfrac{y^{2n}+4y^n+4}{5y^n+25} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{y^{2n}+7y^{n}+10}{10}\cdot\dfrac{5y^n+25}{y^{2n}+4y^n+4} .\end{array} Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(y^n+5)(y^n+2)}{10}\cdot\dfrac{5(y^n+5)}{(y^{n}+2)(y^{n}+2)} \\= \dfrac{(y^n+5)(\cancel{y^n+2})}{\cancel{5}(2)}\cdot\dfrac{\cancel5(y^n+5)}{(\cancel{y^n+2})(y^{n}+2)} \\= \dfrac{(y^n+5)^2}{2(y^{n}+2)} .\end{array}
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