# For the polynomial equation 3x^3-x^2- 7x+8, how do you determine all possible rational roots?

Mar 16, 2017

The only "possible" rational zeros are:

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

#### Explanation:

Given:

$3 {x}^{3} - {x}^{2} - 7 x + 8$

By the rational root theorem, any rational zeros of this cubic are 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 $3$ of the leading term.

That means that the only possible rational zeros are:

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

In fact, none of these work, so this cubic only has irrational and/or complex zeros. Its only real zero is approximately $- 1.79$

graph{3x^3-x^2-7x+8 [-5, 5, -2.52, 15]}