# How do you find all the zeros of  2x^3+9x^2+6x-8?

Aug 13, 2016

This cubic has zeros $- 2$ and $\frac{1}{4} \left(- 5 \pm \sqrt{57}\right)$

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + 9 {x}^{2} + 6 x - 8$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 8$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm 4 , \pm 8$

We find:

$f \left(- 2\right) = 2 \left(- 8\right) + 9 \left(4\right) + 6 \left(- 2\right) - 8 = - 16 + 36 - 12 - 8 = 0$

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

$2 {x}^{3} + 9 {x}^{2} + 6 x - 8 = \left(x + 2\right) \left(2 {x}^{2} + 5 x - 4\right)$

The remaining quadratic is in the form $a {x}^{2} + b x + c$ with $a = 2$, $b = 5$, $c = - 4$.

We can solve this using the quadratic formula:

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

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

$= \frac{1}{4} \left(- 5 \pm \sqrt{25 + 32}\right)$

$= \frac{1}{4} \left(- 5 \pm \sqrt{57}\right)$