# How do you find all the zeros of f(x)=x^4+5x^3-2x^2-18x-12?

Aug 4, 2016

$f \left(x\right)$ has zeros $- 1 , 2 , - 3 - \sqrt{3}$ and $- 3 + \sqrt{3}$

#### Explanation:

$f \left(x\right) = {x}^{4} + 5 {x}^{3} - 2 {x}^{2} - 18 x - 12$

Note that:

$f \left(- 1\right) = 1 - 5 - 2 + 18 - 12 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{4} + 5 {x}^{3} - 2 {x}^{2} - 18 x - 12$

$= \left(x + 1\right) \left({x}^{3} + 4 {x}^{2} - 6 x - 12\right)$

Next try substituting $x = 2$ in the remaining cubic to get:

$8 + 4 \left(4\right) - 6 \left(2\right) - 12 = 8 + 16 - 12 - 12 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{3} + 4 {x}^{2} - 6 x - 12$

$= \left(x - 2\right) \left({x}^{2} + 6 x + 6\right)$

$= \left(x - 2\right) \left({\left(x + 3\right)}^{2} - 9 + 6\right)$

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

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

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

Hence zeros $x = - 3 \pm \sqrt{3}$