# How do you factor u^4-81?

Nov 23, 2016

${u}^{4} - 81 = \left(u - 3\right) \left(u + 3\right) \left({u}^{2} + 9\right)$

#### Explanation:

The difference of squares identity can be written:

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

We can use this a couple of times to derive the factors with Real coefficients, as follows:

${u}^{4} - 81 = {\left({u}^{2}\right)}^{2} - {9}^{2}$

$\textcolor{w h i t e}{{u}^{4} - 81} = \left({u}^{2} - 9\right) \left({u}^{2} + 9\right)$

$\textcolor{w h i t e}{{u}^{4} - 81} = \left({u}^{2} - {3}^{2}\right) \left({u}^{2} + 9\right)$

$\textcolor{w h i t e}{{u}^{4} - 81} = \left(u - 3\right) \left(u + 3\right) \left({u}^{2} + 9\right)$

The remaining quadratic factor has no simpler linear factors with Real coefficients since ${u}^{2} + 9 \ge 9$ for any Real value of $u$, hence no Real zeros or corresponding linear factors.