# Question #2019c

Mar 3, 2017

$x \left(x + 2\right) \left(x - 2\right) = 0$

so $x = - 2 , 0 , 2$

#### Explanation:

You can take out a factor of $x$ because everything has at least one $x$, so

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

Now you have a squared $x$ minus a square number, which is always in the form $\left(x + b\right) \left(x - b\right)$ where $b$ is the root of the square number.

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

So you can say the whole thing is

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

which gives the solutions of $x = 0 , - 2 , + 2$