# How do you factor completely x^4-81?

Aug 14, 2016

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

$\left({x}^{2} + 9\right) \left({x}^{2} - 9\right) = \left({x}^{2} + 9\right) \left(x + 3\right) \left(x - 3\right)$

Aug 14, 2016

$\left(x - 3\right) \left(x + 3\right) \left({x}^{2} + 9\right)$

#### Explanation:

This is a $\textcolor{b l u e}{\text{difference of squares}}$ and, in general, factorises as follows.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$

here ${\left({x}^{2}\right)}^{2} = {x}^{4} \text{ and } {\left(9\right)}^{2} = 81$

$\Rightarrow a = {x}^{2} \text{ and } b = 9$

substituting into (A)

$\Rightarrow {x}^{4} - 81 = \left({x}^{2} - 9\right) \left({x}^{2} + 9\right) \ldots \ldots . . \left(B\right)$

Now, the factor ${x}^{2} - 9 \text{ is also a "color(blue)"difference of squares}$

$\Rightarrow {x}^{2} - 9 = \left(x - 3\right) \left(x + 3\right)$

substituting into (B) to complete the factorising.

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