# How do you factor 9x^3-x?

Jan 25, 2017

$x \left(3 x + 1\right) \left(3 x - 1\right)$

#### Explanation:

$x$ is common to both terms, so you can take it out of both and put the other factor in a bracket.

$9 {x}^{3} - x = x \left(9 {x}^{2} - 1\right)$.

$\left({x}^{2} - 1\right)$ and similar expressions are known as the difference of two squares, and can be further factorised to give:

$x \left(3 x + 1\right) \left(3 x - 1\right)$.

Jan 25, 2017

$x \left(3 x - 1\right) \left(3 x + 1\right)$

#### Explanation:

$9 {x}^{3} - x$

$\therefore = x \left(9 {x}^{2} - 1\right)$

$\therefore = x \left(3 x - 1\right) \left(3 x + 1\right)$

Jan 25, 2017

$x \left(3 x - 1\right) \left(3 x + 1\right)$

#### Explanation:

There is a $\textcolor{b l u e}{\text{common factor}}$ of x in both terms.

$\Rightarrow x \left(9 {x}^{2} - 1\right) \leftarrow \textcolor{red}{\text{remove common factor}}$

$9 {x}^{2} - 1 \text{ is a"color(blue)" difference of squares}$ and factorises, in general as shown.

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

$\text{here " (3x)^2=9x^2" and } {1}^{2} = 1$

$\Rightarrow a = 3 x \text{ and } b = 1$

$\Rightarrow 9 {x}^{2} - 1 = \left(3 x - 1\right) \left(3 x + 1\right)$

$\Rightarrow 9 {x}^{3} - 1 = x \left(3 x - 1\right) \left(3 x + 1\right) \leftarrow \textcolor{red}{\text{ fully factorised}}$