Rewrite $z^{4}-81$ as $\left(z^{2}\right)^{2}-9^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(z^{2}-9\right)\left(z^{2}+9\right)$$
Consider $z^{2}-9$. Rewrite $z^{2}-9$ as $z^{2}-3^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(z-3\right)\left(z+3\right)$$
Rewrite the complete factored expression. Polynomial $z^{2}+9$ is not factored since it does not have any rational roots.
