# How to do this question?

Feb 24, 2018

See below

#### Explanation:

Rational root theorem states that rational roots of a polynomial will take the form of $\frac{p}{q}$ where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. So, for $f \left(x\right) = 12 {x}^{3} + 20 {x}^{2} - x - 6$

$p = \left\{1 , 2 , 3 , 6\right\}$
$q = \left\{1 , 2 , 3 , 4 , 6 , 12\right\}$

$\frac{p}{q} = \left\{\pm 1 , \pm \frac{1}{2} , \pm \frac{1}{3} , \pm \frac{1}{4} , \pm \frac{1}{6} , \pm \frac{1}{12} , \pm 2 , \pm \frac{2}{3} , \pm 3 , \pm \frac{3}{2} , \pm \frac{3}{4} , \pm 6\right\}$ (take each value of p and divide it by each and every value of q)

Now, theoretically, we would test each of these $\frac{p}{q}$ values to find which ones are actually zeros of the function $f \left(x\right) = 12 {x}^{3} + 20 {x}^{2} - x - 6$

Truthfully, what you can do is find the rational zeros in your calculator and prove that they are zeros by using traditional or synthetic substitution. Once you know a root, the factor is $\left(x - a\right)$ where $a$ is your root.