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

Jul 17, 2015

Use difference of squares identity twice to find:

${x}^{4} - 1 = \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right)$

#### Explanation:

The difference of squares identity is:

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

So we find:

${x}^{4} - 1$

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

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

Note that ${x}^{2} + 1$ has no simpler linear factors with real coefficients since ${x}^{2} + 1 \ge 1 > 0$ for all $x \in \mathbb{R}$