# What are the zeros in x^3+3x^2-3x-5?

Jan 7, 2016

$\left\{- 1 , - 1 + \sqrt{6} , - 1 - \sqrt{6}\right\}$

#### Explanation:

Applying the rational roots theorem, we find that any rational roots of the given expression will be of the form $\frac{p}{q}$ where $p$ is a divisor of $- 5$ and $q$ is a positive divisor of $1$. Then, the possible rational roots are $\pm 1$ and $\pm 5$.

Trying these out, we find that ${\left(- 1\right)}^{3} + 3 {\left(- 1\right)}^{2} - 3 \left(- 1\right) - 5 = 0$ Thus $x - \left(- 1\right) = x + 1$ is a factor of ${x}^{3} + 3 {x}^{2} - 3 x - 5$

Dividing, we get
${x}^{3} + {3}^{2} - 3 x - 5 = \left(x + 1\right) \left({x}^{2} + 2 x - 5\right)$

To find the remaining roots, we can simply apply the quadratic formula
$a {x}^{2} + b x + c = 0 \implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
to obtain the roots of ${x}^{2} + 2 x - 5$ as

$\frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(- 5\right)}}{2 \left(1\right)} = - 1 \pm \sqrt{6}$

The the final set of roots (zeros) of the expression is

$\left\{- 1 , - 1 + \sqrt{6} , - 1 - \sqrt{6}\right\}$