# How do you factor completely x^4 - 1?

Sep 11, 2016

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

#### Explanation:

We have: ${x}^{4} - 1$

We can express ${x}^{4}$ as ${\left({x}^{2}\right)}^{2}$:

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

Now that we have a difference of two squares we can factorise in the following way:

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

We now have another difference of two squares.

Let's factorise again to get:

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