# How do you find the rest of the zeros given one of the zero c=-2 and the function f(x)=x^3+2x^2-3x-6?

##### 1 Answer
Nov 30, 2016

$f \left(x\right) = 0$ for $x = - 2$ and $x = \pm \sqrt{3}$

#### Explanation:

If $x = - 2$ is a root, then $f \left(x\right)$ must be divisible by $\left(x + 2\right)$.

We can factor $f \left(x\right)$ using synthetic division :

-2 | 1...... 2....-3.....-6
..............-2......0......6
.......1.......0.....-3......0

The remainder is null, so:

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

Then the other two zeroes are given by:

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

$x = \pm \sqrt{3}$