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

The simplified form of the rational expression $\frac{{{z}^{2}}+2z+1}{8z}\div \frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}$ is $\frac{\left( z+1 \right)\left( z+1 \right)\left( z-1 \right)}{2z\left( z-2 \right)}$.
$\frac{{{z}^{2}}+2z+1}{8z}\div \frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}$ $\frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}$ The reciprocal of $\frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}$ is $\frac{4{{z}^{2}}-4}{{{z}^{2}}-z-2}$ So, multiply the reciprocal of the divisor, \begin{align} & \frac{{{z}^{2}}+2z+1}{8z}\div \frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}=\frac{{{z}^{2}}+2z+1}{8z}\cdot \frac{4{{z}^{2}}-4}{{{z}^{2}}-z-2} \\ & =\frac{\left( {{z}^{2}}+2z+1 \right)\left( 4{{z}^{2}}-4 \right)}{8z\left( {{z}^{2}}-z-2 \right)} \end{align} \begin{align} & \left( {{z}^{2}}+2z+1 \right)\left( 4{{z}^{2}}-4 \right)=\left( {{z}^{2}}+2z+{{1}^{2}} \right)4\left( {{z}^{2}}-1 \right) \\ & =4{{\left( z+1 \right)}^{2}}\left( {{z}^{2}}-{{1}^{2}} \right) \\ & =4\left( z+1 \right)\left( z+1 \right)\left( z+1 \right)\left( z-1 \right) \end{align} \begin{align} & 8z\left( {{z}^{2}}-z-2 \right)=8z\left( {{z}^{2}}-2z+z-2 \right) \\ & =8z\left( z\left( z-2 \right)+\left( z-2 \right) \right) \\ & =8z\left( z+1 \right)\left( z-2 \right) \end{align} So, the rational expression becomes, $\frac{{{z}^{2}}+2z+1}{8z}\div \frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}=\frac{4\left( z+1 \right)\left( z+1 \right)\left( z+1 \right)\left( z-1 \right)}{8z\left( z+1 \right)\left( z-2 \right)}$ Regroup and remove the factor equal to 1, \begin{align} & \frac{{{z}^{2}}+2z+1}{8z}\div \frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}=\frac{\left( z+1 \right)\left( z+1 \right)4\left( z+1 \right)\left( z-1 \right)}{8z\left( z+1 \right)\left( z-2 \right)} \\ & =\frac{\left( z+1 \right)}{\left( z+1 \right)}\cdot \frac{4}{4}\cdot \frac{\left( z+1 \right)\left( z+1 \right)\left( z-1 \right)}{2z\left( z-2 \right)} \\ & =1\cdot 1\cdot \frac{\left( z+1 \right)\left( z+1 \right)\left( z-1 \right)}{2z\left( z-2 \right)} \\ & =\frac{{{\left( z+1 \right)}^{2}}\left( z-1 \right)}{2z\left( z-2 \right)} \end{align}