# How do you write (z^3-8)/(z^2+2z+4) in simplest form?

$\left(\frac{\left(z - 2\right) \left({z}^{2} + 2 z + 4\right)}{\left({z}^{2} + 2 z + 4\right)}\right)$
$\frac{\left({z}^{2} + 2 z + 4\right)}{\left({z}^{2} + 2 z + 4\right)} = 1$
Giving us $\left(z - 2\right) \cdot 1$ which is
$z - 2$