# How do you factor completely w^2 - w^2 y^4?

##### 1 Answer
Apr 4, 2018

${w}^{2} \left(1 + y\right) \left(1 - y\right) \left(1 + {y}^{2}\right)$

#### Explanation:

This is just a difference of squares. When you have ${a}^{2} - {b}^{2}$, it's is equal to $\left(a + b\right) \left(a - b\right)$. Just simply expand it to see why. In this case, it would be

${\left(w\right)}^{2} - {\left(w \cdot {y}^{2}\right)}^{2}$

$\left(w - w {y}^{2}\right) \left(w + w {y}^{2}\right)$

${w}^{2} \left(1 - {y}^{2}\right) \left(1 + {y}^{2}\right)$

${w}^{2} \left(1 + y\right) \left(1 - y\right) \left(1 + {y}^{2}\right)$