# How do you factor 128x^3 - 1024?

Feb 6, 2017

$128 \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

#### Explanation:

The 2 terms have a $\textcolor{b l u e}{\text{common factor}}$ of 128, which can be taken out.

$\Rightarrow 128 \left({x}^{3} - 8\right) \to \left(A\right)$

${x}^{3} - 8$ is a $\textcolor{b l u e}{\text{difference of cubes}}$ which, in general, is factorised as shown.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

using a = x and b = 2, then

$\Rightarrow {x}^{3} - 8 = \left(x - 2\right) \left({x}^{2} + 2 x + {2}^{2}\right) = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

Returning to (A) the complete factorisation is.

$128 {x}^{3} - 1024 = 128 \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$