# How do you use the rational roots theorem to find all possible zeros of f(x)=3x^3-17x^2+18x+8?

Jun 5, 2016

$x = - \frac{1}{3}$, $x = 2$, $x = 4$

#### Explanation:

$f \left(x\right) = 3 {x}^{3} - 17 {x}^{2} + 18 x + 8$

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be 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:

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

We find:

$f \left(- \frac{1}{3}\right) = - \frac{1}{9} - \frac{17}{9} - 6 + 8 = 0$

So $x = - \frac{1}{3}$ is a zero and $\left(3 x + 1\right)$ a factor of $f \left(x\right)$

$3 {x}^{3} - 17 {x}^{2} + 18 x + 8$

$= \left(3 x + 1\right) \left({x}^{2} - 6 x + 8\right)$

$= \left(3 x + 1\right) \left(x - 2\right) \left(x - 4\right)$

Hence the other two zeros are $x = 2$ and $x = 4$