## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter R - Algebra Reference - R.3 Rational Expressions - R.3 Exercises - Page R-11: 24

#### Answer

$\dfrac{4n^{2}+4n-3}{6n^{2}-n-15}\cdot\dfrac{8n^{2}+32n+30}{4n^{2}+16n+15}=\dfrac{2(2n-1)}{3n-5}$

#### Work Step by Step

$\dfrac{4n^{2}+4n-3}{6n^{2}-n-15}\cdot\dfrac{8n^{2}+32n+30}{4n^{2}+16n+15}$ Factor both rational expressions completely: $\dfrac{4n^{2}+4n-3}{6n^{2}-n-15}\cdot\dfrac{8n^{2}+32n+30}{4n^{2}+16n+15}=...$ $...=\dfrac{(2n+3)(2n-1)}{(2n+3)(3n-5)}\cdot\dfrac{2(2n+5)(2n+3)}{(2n+5)(2n+3)}=...$ Evaluate the product of the two rational expressions: $...=\dfrac{2(2n+3)^{2}(2n-1)(2n+5)}{(2n+3)^{2}(3n-5)(2n+5)}=...$ Simplify by removing the factors that appear both in the numerator and in the denominator of the resulting rational expression: $...=\dfrac{2(2n-1)}{3n-5}$

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