# How do you find all the zeros of x^3+2x^2-2x-3?

Feb 27, 2016

Use the rational root theorem to help find the first zero $x = - 1$, then divide $f \left(x\right)$ by $\left(x + 1\right)$ to find a quadratic and hence the other two zeros:

$x = \frac{- 1 \pm \sqrt{13}}{2}$

#### Explanation:

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

By the rational root theorem, any rational zeros of $f \left(x\right)$ will be expressible in the form $\frac{p}{q}$ for some integers $p$ and $q$ with $p$ a divisor of the constant term $- 3$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 3$

Trying each in turn, we find:

$f \left(1\right) = 1 + 2 - 2 - 3 = - 2$

$f \left(- 1\right) = - 1 + 2 + 2 - 3 = 0$

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

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

The remaining quadratic factor is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 1$ and $c = - 3$. This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 1 \pm \sqrt{1 - \left(4 \cdot 1 \cdot - 3\right)}}{2 \cdot 1}$

$= \frac{- 1 \pm \sqrt{13}}{2}$