# How do you use the rational roots theorem to find all possible zeros of P(x)=2x^3+9x^2+11x-8?

Aug 12, 2016

$P \left(x\right)$ has zeros $\frac{1}{2}$ and $\frac{5}{2} \pm \frac{\sqrt{7}}{2} i$

#### Explanation:

$P \left(x\right) = 2 {x}^{3} + 9 {x}^{2} + 11 x - 8$

By the rational roots theorem, any rational zeros of $P \left(x\right)$ 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:

$P \left(\frac{1}{2}\right) = 2 \left(\frac{1}{8}\right) + 9 \left(\frac{1}{4}\right) + 11 \left(\frac{1}{2}\right) - 8 = \frac{1 + 9 + 22 - 32}{4} = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$2 {x}^{3} + 9 {x}^{2} + 11 x - 8$

$= \left(2 x - 1\right) \left({x}^{2} + 5 x + 8\right)$

$= \left(2 x - 1\right) \left({\left(x - \frac{5}{2}\right)}^{2} - \frac{25}{4} + 8\right)$

$= \left(2 x - 1\right) \left({\left(x - \frac{5}{2}\right)}^{2} + \frac{7}{4}\right)$

$= \left(2 x - 1\right) \left({\left(x - \frac{5}{2}\right)}^{2} - {\left(\frac{\sqrt{7}}{2} i\right)}^{2}\right)$

$= \left(2 x - 1\right) \left(x - \frac{5}{2} - \frac{\sqrt{7}}{2} i\right) \left(x - \frac{5}{2} + \frac{\sqrt{7}}{2} i\right)$

Hence the other two zeros are:

$x = \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$