## Precalculus (6th Edition) Blitzer

The required solution is $\frac{1}{6}$.
We have the given algebraic expression: $\frac{1}{{{x}^{2}}-2x-8}\div \left( \frac{1}{x-4}-\frac{1}{x+2} \right)$. For an algebraic expression, a rational expression is an expression that can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator $q\ne 0$. Now, solve the bracket of the above rational expression: \begin{align} & \frac{1}{x-4}-\frac{1}{x+2}=\frac{1}{x-4}\times \frac{\left( x+2 \right)}{\left( x+2 \right)}-\frac{1}{x+2}\times \frac{\left( x-4 \right)}{\left( x-4 \right)} \\ & =\frac{\left( x+2 \right)}{\left( x+2 \right)\left( x-4 \right)}-\frac{\left( x-4 \right)}{\left( x+2 \right)\left( x-4 \right)} \\ & =\frac{{x}+2-{x}+4}{\left( x+2 \right)\left( x-4 \right)} \\ & =\frac{6}{\left( x+2 \right)\left( x-4 \right)} \end{align}. So, the given expression becomes $\frac{1}{{{x}^{2}}-2x-8}\div \left( \frac{1}{x-4}-\frac{1}{x+2} \right)=\frac{1}{{{x}^{2}}-2x-8}\div \frac{6}{\left( x+2 \right)\left( x-4 \right)}$. And factorize ${{x}^{2}}-2x-8$ by splitting the middle term: \begin{align} & {{x}^{2}}-2x-8={{x}^{2}}+\left( -4+2 \right)x-8 \\ & ={{x}^{2}}-4x+2x-8 \\ & =x\left( x-4 \right)+2\left( x-4 \right) \\ & =\left( x-4 \right)\left( x+2 \right) \end{align}. Therefore, the given expression gets reduced to \begin{align} & \frac{1}{{{x}^{2}}-2x-8}\div \left( \frac{1}{x-4}-\frac{1}{x+2} \right)=\frac{1}{\left( x-4 \right)\left( x+2 \right)}\div \frac{6}{\left( x+2 \right)\left( x-4 \right)} \\ & =\frac{1}{\left( x-4 \right)\left( x+2 \right)}\times \frac{\left( x+2 \right)\left( x-4 \right)}{6} \\ & =\frac{1}{\left( x+2 \right)\left( x-4 \right)}\times \frac{\left( x+2 \right)\left( x-4 \right)}{6} \\ & =\frac{1}{6} \end{align}. Hence, $\frac{1}{{{x}^{2}}-2x-8}\div \left( \frac{1}{x-4}-\frac{1}{x+2} \right)=$ $\frac{1}{6}$.