# How to do I solve this problem? 3x^2-12=0

## I can't remember how to do these and just need step by step so I can finish the problem. This is only a portion of my calculus problem but I forget how to do this part. The whole problem is f(x) = x^3-12x+17 find the relative extreme points of the function if they exist. I did the first part and found the derivatives of the problem but now I am stuck at the next step which is what I posted for my question. It's been years since I took algebra so I forget. Thanks!

Feb 3, 2017

$x = - 2 , x = 2$

#### Explanation:

You need to factor $3 {x}^{2} - 12$, which ends up being

$3 \left(x + 2\right) \left(x - 2\right)$

and then set each of these (but ignore the $3$) equal to zero and solve.

$x + 2 = 0$

$x + 2 - 2 = 0 - 2$

$x = - 2$

and

$x - 2 = 0$

$x - 2 + 2 = 0 + 2$

$x = 2$

So $x = 2 , - 2$