# How do you factor 9x^4y^2 - z^6?

Mar 24, 2018

See a solution process below:

#### Explanation:

This is a special case of a quadratic:

${\textcolor{red}{a}}^{2} - {\textcolor{b l u e}{b}}^{2} = \left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right) \left(\textcolor{red}{a} - \textcolor{b l u e}{b}\right)$

• Let $\textcolor{red}{a} = 3 {x}^{2} y$; Then ${\textcolor{red}{a}}^{2} = 9 {x}^{4} {y}^{2}$

• Let $\textcolor{b l u e}{b} = {z}^{3}$; Then ${\textcolor{b l u e}{b}}^{2} = {z}^{6}$

Substituting gives:

$\textcolor{red}{9 {x}^{4} {y}^{2}} - \textcolor{b l u e}{{z}^{6}} \implies$

${\textcolor{red}{\left(3 {x}^{2} y\right)}}^{2} - {\textcolor{b l u e}{\left({z}^{3}\right)}}^{2} \implies$

$\left(\textcolor{red}{3 {x}^{2} y} + \textcolor{b l u e}{{z}^{3}}\right) \left(\textcolor{red}{3 {x}^{2} y} - \textcolor{b l u e}{{z}^{3}}\right)$