# How do you factor 4x^4 - 64 ?

Sep 3, 2016

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

#### Explanation:

The first step is to take out a $\textcolor{b l u e}{\text{common factor}}$ of 4

$\Rightarrow 4 \left({x}^{4} - 16\right) \ldots \ldots . . \left(A\right)$

Now ${x}^{4} - 16 \text{ is a " color(blue)"difference of squares}$ and in general factorises as.

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

${x}^{4} = {\left({x}^{2}\right)}^{2} \text{ and " 16=(4)^2rArra=x^2" and } b = 4$

substitute these values into (B)

$\Rightarrow {x}^{4} - 16 = \left({x}^{2} - 4\right) \left({x}^{2} + 4\right)$

Note that ${x}^{2} - 4$ is also a $\textcolor{b l u e}{\text{difference of squares}}$

and using the method of above

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

substituting the values from the 2 results into (A) remembering the factor $\left({x}^{2} + 4\right)$

$\Rightarrow 4 {x}^{4} - 64 = 4 \left(x - 2\right) \left(x + 2\right) \left({x}^{2} + 4\right)$