# How do you use the rational roots theorem to find all possible zeros of f(x)=x^3-5x^2+2x+12?

Jul 24, 2016

$x = 3$ or $x = 1 \pm \sqrt{5}$

#### Explanation:

$f \left(x\right) = {x}^{3} - 5 {x}^{2} + 2 x + 12$

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 $12$ 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 find:

$f \left(3\right) = 27 - 45 + 6 + 12 = 0$

So $x = 3$ is a zero and $\left(x - 3\right)$ a factor:

${x}^{3} - 5 {x}^{2} + 2 x + 12$

$= \left(x - 3\right) \left({x}^{2} - 2 x - 4\right)$

$= \left(x - 3\right) \left({x}^{2} - 2 x + 1 - 5\right)$

$= \left(x - 3\right) \left({\left(x - 1\right)}^{2} - {\left(\sqrt{5}\right)}^{2}\right)$

$= \left(x - 3\right) \left(x - 1 - \sqrt{5}\right) \left(x - 1 + \sqrt{5}\right)$

So the remaining two zeros are:

$x = 1 \pm \sqrt{5}$