## Intermediate Algebra (6th Edition)

$\left(\dfrac{1}{3}-2z\right)\left(\dfrac{1}{3}+2z\right)$
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then, \begin{align*} \dfrac{1}{9}-4z^2 \Rightarrow \left(\dfrac{1}{3}-2z\right)\left(\dfrac{1}{3}+2z\right) .\end{align*}