# What are all the possible rational zeros for f(x)=6x^3-x^2-13x+8 and how do you find all zeros?

Dec 24, 2016

This cubic has "possible" rational zeros:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm \frac{8}{3} , \pm 4 , \pm 8$

and actual zeros:

$1 \text{ }$ and $\text{ } - \frac{5}{12} \pm \frac{\sqrt{217}}{12}$

#### Explanation:

$f \left(x\right) = 6 {x}^{3} - {x}^{2} - 13 x + 8$

By the rational roots theorem, any rational zero of $f \left(x\right)$ is 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 $6$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm \frac{8}{3} , \pm 4 , \pm 8$

There is a shortcut in that the sum of the coefficients is $0$.

That is:

$6 - 1 - 13 + 8 = 0$

So $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

$6 {x}^{3} - {x}^{2} - 13 x + 8 = \left(x - 1\right) \left(6 {x}^{2} + 5 x - 8\right)$

The remaining quadratic factor is in the form $a {x}^{2} + b x + c$ with $a = 6$, $b = 5$ and $c = - 8$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {5}^{2} - 4 \left(6\right) \left(- 8\right) = 25 + 192 = 217$

Since $\Delta > 0$ this quadratic has two Real zeros, but $\Delta$ is not a perfect square, so those zeros are irrational.

We can find the zeros using the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- 5 \pm \sqrt{217}}{12}$

$\textcolor{w h i t e}{x} = - \frac{5}{12} \pm \frac{\sqrt{217}}{12}$