# How do you use the rational root theorem to find the roots of x^4-x^3-2x^2-4x-24?

Sep 6, 2015

Use rational root theorem to find possible rational zeros. Try the first few, then use synthetic division with the roots found to find there are no more Real ones.

$x = - 2$ or $x = 3$

#### Explanation:

Let $f \left(x\right) = {x}^{4} - {x}^{3} - 2 {x}^{2} - 4 x - 24$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be expressible in the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $q > 0$, $p$ a divisor of the constant term $- 24$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 8$, $\pm 12$ and $\pm 24$.

Let us try a few:

$f \left(1\right) = 1 - 1 - 2 - 4 - 24 = - 30$
$f \left(- 1\right) = 1 + 1 - 2 + 4 - 24 = - 20$
$f \left(2\right) = 16 - 8 - 8 - 8 - 24 = - 32$
$\textcolor{b l u e}{f \left(- 2\right) = 16 + 8 - 8 + 8 - 24 = 0}$
$\textcolor{b l u e}{f \left(3\right) = 81 - 27 - 18 - 12 - 24 = 0}$

So $- 2$ and $3$ are roots. Before going further, let's divide $f \left(x\right)$ by the corresponding factors $\left(x + 2\right)$ and $\left(x - 3\right)$

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

Use synthetic division:

So ${x}^{4} - {x}^{3} - 2 {x}^{2} - 4 x - 24 = \left({x}^{2} - x - 6\right) \left({x}^{2} + 4\right)$

${x}^{2} + 4$ has no Real zeros since ${x}^{2} \ge 0$ for all $x \in \mathbb{R}$