# What are all the possible rational zeros for f(x)=5x^3-9x^2-12x-2 and how do you find all zeros?

Jun 17, 2017

The zeros of $f \left(x\right)$ are:

$- \frac{1}{5} \text{ }$ and $\text{ } 1 \pm \sqrt{3}$

#### Explanation:

Given:

$f \left(x\right) = 5 {x}^{3} - 9 {x}^{2} - 12 x - 2$

By the rational root 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 $- 2$ and $q$ a divisor of the coefficient $5$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{5} , \pm \frac{2}{5} , \pm 1 , \pm 2$

First check $f \left(\pm 1\right)$...

$f \left(1\right) = 5 - 9 - 12 - 2 = - 18$

$f \left(- 1\right) = - 5 - 9 + 12 - 2 = - 4$

We can deduce that not all of the zeros are rational, since there are not enough factors to go around.

Trying the other possibilities, we eventually find:

$f \left(- \frac{1}{5}\right) = - 5 \left(\frac{1}{125}\right) + 9 \left(\frac{1}{25}\right) + 12 \left(\frac{1}{5}\right) - 2$

$\textcolor{w h i t e}{f \left(- \frac{1}{5}\right)} = \frac{- 1 - 9 + 60 - 50}{25} = 0$

So $x = - \frac{1}{5}$ is a zero and $\left(5 x + 1\right)$ a factor:

$5 {x}^{3} - 9 {x}^{2} - 12 x - 2 = \left(5 x + 1\right) \left({x}^{2} - 2 x - 2\right)$

$\textcolor{w h i t e}{5 {x}^{3} - 9 {x}^{2} - 12 x - 2} = \left(5 x + 1\right) \left({x}^{2} - 2 x + 1 - 3\right)$

$\textcolor{w h i t e}{5 {x}^{3} - 9 {x}^{2} - 12 x - 2} = \left(5 x + 1\right) \left({\left(x - 1\right)}^{2} - {\left(\sqrt{3}\right)}^{2}\right)$

$\textcolor{w h i t e}{5 {x}^{3} - 9 {x}^{2} - 12 x - 2} = \left(5 x + 1\right) \left(x - 1 - \sqrt{3}\right) \left(x - 1 + \sqrt{3}\right)$

So the zeros are:

$- \frac{1}{5} \text{ }$ and $\text{ } 1 \pm \sqrt{3}$