# How do you factor 2x^3 -128x?

Oct 18, 2015

$2 x \left(x - 8\right) \left(x + 8\right)$

#### Explanation:

Start by breaking down your initial expression in search of possible common factors

$2 {x}^{3} = \textcolor{b l u e}{2 \cdot x} \cdot {x}^{2}$

$128 x = 64 \cdot \textcolor{b l u e}{2 \cdot x}$

This means that you can use $\textcolor{b l u e}{2 x}$ as a common factor for these two terms

$2 {x}^{3} - 128 x = 2 x \cdot \left({x}^{2} - 64\right)$

Notice that $64$ is a perfect square

$64 = 8 \cdot 8 = {8}^{2}$

which means that you are actually dealing with the difference of two squares

$\textcolor{b l u e}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$

The bracket can thus be written as

${x}^{2} - 64 = {x}^{2} - {8}^{2} = \left(x - 8\right) \left(x + 8\right)$

The expression will thus be equivalent to

$2 {x}^{3} - 128 x = \textcolor{g r e e n}{2 x \left(x - 8\right) \left(x + 8\right)}$