Consider $zx^{3}+2x^{3}-8z-16$. Do the grouping $zx^{3}+2x^{3}-8z-16=\left(zx^{3}+2x^{3}\right)+\left(-8z-16\right)$, and factor out $x^{3}$ in the first and $-8$ in the second group.
$$x^{3}\left(z+2\right)-8\left(z+2\right)$$
Factor out common term $z+2$ by using distributive property.
$$\left(z+2\right)\left(x^{3}-8\right)$$
Consider $x^{3}-8$. Rewrite $x^{3}-8$ as $x^{3}-2^{3}$. The difference of cubes can be factored using the rule: $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$.
$$\left(x-2\right)\left(x^{2}+2x+4\right)$$
Rewrite the complete factored expression. Polynomial $x^{2}+2x+4$ is not factored since it does not have any rational roots.
