# Using the rational root theorem, what are the possible rational roots of x^3-34x+12=0 ?

Oct 6, 2016

According to the theorem, the possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 12$

#### Explanation:

$f \left(x\right) = {x}^{3} - 34 x + 12$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integeres $p , q$ with $p$ a divisor of the constant term $12$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 12$

Trying each in turn, we eventually find that:

$f \left(\textcolor{b l u e}{- 6}\right) = {\left(\textcolor{b l u e}{- 6}\right)}^{3} - 34 \left(\textcolor{b l u e}{- 6}\right) + 12$

$\textcolor{w h i t e}{f \left(\textcolor{w h i t e}{- 6}\right)} = - 216 + 204 + 12$

$\textcolor{w h i t e}{f \left(\textcolor{w h i t e}{- 6}\right)} = 0$

So $x = - 6$ is a rational root.

The other two roots are Real but irrational.