Answer
The simplified form of the rational expression $\frac{{{z}^{2}}+2z+1}{8z}\div \frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}$ is $\frac{\left( z+1 \right)\left( z+1 \right)\left( z-1 \right)}{2z\left( z-2 \right)}$.
Work Step by Step
$\frac{{{z}^{2}}+2z+1}{8z}\div \frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}$
$\frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}$
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,
$\begin{align}
& \frac{{{z}^{2}}+2z+1}{8z}\div \frac{{{z}^{2}}-z-2}{4{{z}^{2}}-4}=\frac{\left( z+1 \right)\left( z+1 \right)4\left( z+1 \right)\left( z-1 \right)}{8z\left( z+1 \right)\left( z-2 \right)} \\
& =\frac{\left( z+1 \right)}{\left( z+1 \right)}\cdot \frac{4}{4}\cdot \frac{\left( z+1 \right)\left( z+1 \right)\left( z-1 \right)}{2z\left( z-2 \right)} \\
& =1\cdot 1\cdot \frac{\left( z+1 \right)\left( z+1 \right)\left( z-1 \right)}{2z\left( z-2 \right)} \\
& =\frac{{{\left( z+1 \right)}^{2}}\left( z-1 \right)}{2z\left( z-2 \right)}
\end{align}$
