Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Test: 14

Answer

The required value is $\frac{2\left( x+3 \right)}{\left( x+1 \right)}$ , $x\ne -1,-4,3,-3$.

Work Step by Step

Consider the expression, $\frac{2x+8}{x-3}\div \frac{{{x}^{2}}+5x+4}{{{x}^{2}}-9}$ The given expression can be further evaluated by inverting the divisor and multiplying: $\begin{align} & \frac{2x+8}{x-3}\div \frac{{{x}^{2}}+5x+4}{{{x}^{2}}-9}=\frac{2x+8}{x-3}\cdot \frac{{{x}^{2}}-9}{{{x}^{2}}+5x+4} \\ & =\frac{2\left( x+4 \right)}{\left( x-3 \right)}\cdot \frac{\left( x+3 \right)\left( x-3 \right)}{{{x}^{2}}+4x+x+4} \\ & =\frac{2\left( x+4 \right)}{\left( x-3 \right)}\cdot \frac{\left( x+3 \right)\left( x-3 \right)}{x\left( x+4 \right)+1\left( x+4 \right)} \\ & =\frac{2\left( x+4 \right)}{\left( x-3 \right)}\cdot \frac{\left( x+3 \right)\left( x-3 \right)}{\left( x+4 \right)\left( x+1 \right)} \end{align}$ Solve further: $\begin{align} & \frac{2x+8}{x-3}\div \frac{{{x}^{2}}+5x+4}{{{x}^{2}}-9}=\frac{2\left( x+3 \right)\left( x+4 \right)\left( x-3 \right)}{\left( x+1 \right)\left( x+4 \right)\left( x-3 \right)} \\ & =\frac{2\left( x+3 \right)}{\left( x+1 \right)} \end{align}$ Here, $x\ne -1,-4,3,-3$ because at $x=-1,-4,3,-3$ the denominator will become zero. Therefore, the value of $\frac{2x+8}{x-3}\div \frac{{{x}^{2}}+5x+4}{{{x}^{2}}-9}$ is $\frac{2\left( x+3 \right)}{\left( x+1 \right)}$.
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