# How do you use the rational root theorem to list all possible roots for 12x^4+14x^3-5x^2-14x-4=0?

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Aug 30, 2017

The only possible rational roots are:

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

...but none of these is a root.

So this equation has no rational roots.

#### Explanation:

Given:

$12 {x}^{4} + 14 {x}^{3} - 5 {x}^{2} - 14 x - 4 = 0$

By the rational root theorem, any rational roots must be 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 $12$ of the leading term.

So the only possible rational roots are:

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

In practice, none of these is a root, so this quartic equation has no rational roots.

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