# How many zeroes does f(x) = x^3 + 3x^2 - x - 3 have?

Nov 5, 2015

The zeroes are -3, 1, and -1.

#### Explanation:

In order to figure this out, you have to set the equation to 0.

$f \left(x\right) = {x}^{3} + 3 {x}^{2} - x - 3 = 0$

Now looking at it, we could either factor it (which is usually the best way to do it) or use synthetic division (which is probably the best thing to do here). Now I'm going to go through the process of synthetic division on my own assuming you know how to do it, if you don't it's a great method of finding out roots of functions and getting their "remains." I'm sure there's great tutorials on the internet about it. Also it's nearly impossible to format on here in my opinion.

So using synthetic division we get -3 as a zero and ${x}^{2} - 1$ left over.

Now we factor ${x}^{2} - 1$ and get $\left(x - 1\right) \left(x + 1\right)$ which means our other two zeroes are 1 and -1 once we solve $x - 1 = 0$ and $x + 1 = 0$.