# How do you factor  x^3 - 9x = 0?

Feb 20, 2016

$x \left(x + 3\right) \left(x - 3\right) = 0$

#### Explanation:

First, notice that the term $x$ is common in both terms on the left hand side. Thus, it can be factored from the two.

$x \left({x}^{2} - 9\right) = 0$

Now, we should consider the quadratic term ${x}^{2} - 9$. This is actually a difference of squares, since both ${x}^{2}$ and $9$ are squared terms being subtracted: ${x}^{2} = {\left(x\right)}^{2} , 9 = {\left(3\right)}^{2}$.

Differences of squares factor as follows:

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

Hence in ${x}^{2} - 9$ we have $a = x$ and $b = 3$, so the original expression factors into:

$x \left(x + 3\right) \left(x - 3\right) = 0$

This is completely factored.