# How do you use the rational root theorem to list all possible roots for 36x^3+144x^2-x-4=0?

Nov 7, 2016

$\left(x + 4\right) \left(6 x - 1\right) \left(6 x + 1\right) = 0$
Rational roots:
$x = - 4 , x = \pm \frac{1}{6}$

Nov 9, 2016

The rational root theorem gives you possible rational roots:

$\pm \frac{1}{36} , \pm \frac{1}{18} , \pm \frac{1}{12} , \pm \frac{1}{9} , \pm \frac{1}{6} , \pm \frac{2}{9} , \pm \frac{1}{4} , \pm \frac{1}{3} , \pm \frac{4}{9} , \pm \frac{1}{2} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm 4$

The actual roots are:

$\pm \frac{1}{6}$, $- 4$

#### Explanation:

Given:

$f \left(x\right) = 36 {x}^{3} + 144 {x}^{2} - x - 4$

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 $- 4$ and $q$ a divisor of the coefficient $36$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{36} , \pm \frac{1}{18} , \pm \frac{1}{12} , \pm \frac{1}{9} , \pm \frac{1}{6} , \pm \frac{2}{9} , \pm \frac{1}{4} , \pm \frac{1}{3} , \pm \frac{4}{9} , \pm \frac{1}{2} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm 4$

Note well that the rational root theorem will not identify possible irrational or Complex zeros.

In this particular example, we do not have to try each of these rational possibilities, since there is a shortcut...

Note that the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

$0 = 36 {x}^{3} + 144 {x}^{2} - x - 4$

$\textcolor{w h i t e}{0} = \left(36 {x}^{3} + 144 {x}^{2}\right) - \left(x + 4\right)$

$\textcolor{w h i t e}{0} = 36 {x}^{2} \left(x + 4\right) - 1 \left(x + 4\right)$

$\textcolor{w h i t e}{0} = \left(36 {x}^{2} - 1\right) \left(x + 4\right)$

$\textcolor{w h i t e}{0} = \left({\left(6 x\right)}^{2} - {1}^{2}\right) \left(x + 4\right)$

$\textcolor{w h i t e}{0} = \left(6 x - 1\right) \left(6 x + 1\right) \left(x + 4\right)$

Hence roots:

$\pm \frac{1}{6}$ and $- 4$