# How do you factor 7x^2y^2-63y^4 ?

Aug 23, 2017

$7 {x}^{2} {y}^{2} - 63 {y}^{4} = 7 {y}^{2} \left(x - 3 y\right) \left(x + 3 y\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We will use this with $a = {x}^{2}$ and $b = {\left(3 y\right)}^{2}$, but first separate out the common factor $7 {y}^{2}$...

$7 {x}^{2} {y}^{2} - 63 {y}^{4} = 7 {y}^{2} \left({x}^{2} - 9 {y}^{2}\right)$

$\textcolor{w h i t e}{7 {x}^{2} {y}^{2} - 63 {y}^{4}} = 7 {y}^{2} \left({x}^{2} - {\left(3 y\right)}^{2}\right)$

$\textcolor{w h i t e}{7 {x}^{2} {y}^{2} - 63 {y}^{4}} = 7 {y}^{2} \left(x - 3 y\right) \left(x + 3 y\right)$