Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.6 Rational Expressions and Equations - R.6 Exercise Set - Page 980: 20


The simplified form of the rational expression $\left( {{x}^{2}}-16 \right)\div \frac{4x+16}{3{{x}^{2}}}$ is$\frac{3{{x}^{2}}\cdot \left( x-4 \right)}{4}$.

Work Step by Step

$\left( {{x}^{2}}-16 \right)\div \frac{4x+16}{3{{x}^{2}}}$ $\frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}$ The reciprocal of $\frac{4x+16}{3{{x}^{2}}}$ is$\frac{3{{x}^{2}}}{4x+16}$. So, multiply the reciprocal of the divisor, $\begin{align} & \left( {{x}^{2}}-16 \right)\div \frac{4x+16}{3{{x}^{2}}}=\left( {{x}^{2}}-16 \right)\cdot \frac{3{{x}^{2}}}{4x+16} \\ & =\frac{\left( {{x}^{2}}-16 \right)\left( 3{{x}^{2}} \right)}{4x+16} \end{align}$ Factor the numerator: $\begin{align} & \left( {{x}^{2}}-16 \right)\left( 3{{x}^{2}} \right)=\left( {{x}^{2}}-{{4}^{2}} \right)\left( 3{{x}^{2}} \right) \\ & =\left( x+4 \right)\left( x-4 \right)\left( 3{{x}^{2}} \right) \end{align}$ Factor the denominator as: $4x+16=4\left( x+4 \right)$ So, the rational expression becomes, $\left( {{x}^{2}}-16 \right)\div \frac{4x+16}{3{{x}^{2}}}=\frac{\left( x+4 \right)\left( x-4 \right)\left( 3{{x}^{2}} \right)}{4\left( x+4 \right)}$ Regroup and remove the factor equal to 1, $\begin{align} & \left( {{x}^{2}}-16 \right)\div \frac{4x+16}{3{{x}^{2}}}=\frac{\left( x+4 \right)\left( x-4 \right)\left( 3{{x}^{2}} \right)}{\left( x+4 \right)\left( 4 \right)} \\ & =\frac{\left( x+4 \right)}{\left( x+4 \right)}\cdot \frac{\left( x-4 \right)\left( 3{{x}^{2}} \right)}{4} \\ & =1\cdot \frac{\left( 3{{x}^{2}} \right)\left( x-4 \right)}{\left( 4 \right)} \\ & =\frac{3{{x}^{2}}\cdot \left( x-4 \right)}{4} \end{align}$
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