# How do you factor x^3-49x?

Mar 3, 2018

Since we have ${x}^{3}$, we know that we need one $x$ out and two $x$ inside the parentheses.

And we know $\left(7\right) \left(7\right)$ is 49. Since it is -49, one of the seven will be negative.

Thus

The answer is: $x \left(x + 7\right) \left(x - 7\right)$

Mar 3, 2018

$x \left(x + 7\right) \left(x - 7\right)$

#### Explanation:

We can see that both terms contain an $x$ which we can factor out to get $x \left({x}^{2} - 49\right)$

Now we can use difference of two squares to factorise ${x}^{2} - 49$. Difference of two squares tells us that ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

${x}^{2} - 49 = \left(x + 7\right) \left(x - 7\right)$ since ${7}^{2} = 49$

Substituting this gives us $x \left(x + 7\right) \left(x - 7\right)$