# How do you factor 5y^6-5y^2?

##### 1 Answer
Jun 28, 2015

First separate the common factor $5 {y}^{2}$ then factor the remaining well known quartic...

$5 {y}^{6} - 5 {y}^{2}$

$= 5 {y}^{2} \left({y}^{4} - 1\right)$

$= 5 {y}^{2} \left(y - 1\right) \left(y + 1\right) \left({y}^{2} + 1\right)$

#### Explanation:

The difference of squares identity is:

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

We use this twice below...

First separate the common factor $5 {y}^{2}$ then factor the remaining quartic:

$5 {y}^{6} - 5 {y}^{2}$

$= 5 {y}^{2} \left({y}^{4} - 1\right)$

The quartic factor is a difference of squares...

$= 5 {y}^{2} \left({\left({y}^{2}\right)}^{2} - {1}^{2}\right)$

$= 5 {y}^{2} \left({y}^{2} - 1\right) \left({y}^{2} + 1\right)$

The first quadratic factor is a difference of squares...

$= 5 {y}^{2} \left({y}^{2} - {1}^{2}\right) \left({y}^{2} + 1\right)$

$= 5 {y}^{2} \left(y - 1\right) \left(y + 1\right) \left({y}^{2} + 1\right)$

The remaining quadratic factor has no linear factors with real coefficients since ${y}^{2} + 1 \ge 1 > 0$ for all $y \in \mathbb{R}$