# How do you use the rational root theorem to find the roots of  f(x) = x^3-4x^2-2x+8?

Dec 6, 2015

Use the rational root theorem to identify the possible rational roots to try. Hence find $x = 4$ and hence the other two irrational roots $x = \pm \sqrt{2}$

#### Explanation:

$f \left(x\right) = {x}^{3} - 4 {x}^{2} - 2 x + 8$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be expressible in lowest terms in the form $\frac{p}{q}$ for some integers $p$, $q$ where $p$ is a divisor of the constant term $8$ and $q$ is a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$

$\pm 1$ won't work since the leading term would be odd and the remaining terms are even.

$f \left(2\right) = 8 - 16 - 4 + 8 = - 4$

$f \left(- 2\right) = - 8 - 16 + 4 + 8 = - 12$

$f \left(4\right) = 64 - 64 - 8 + 8 = 0$

So $x = 4$ is a root and $\left(x - 4\right)$ a factor.

${x}^{3} - 4 {x}^{2} - 2 x + 8 = \left(x - 4\right) \left({x}^{2} - 2\right)$

The remaining roots are the irrational roots of ${x}^{2} - 2 = 0$, that is:

$x = \pm \sqrt{2}$