# How do you use the rational root theorem to find the roots of x^3 – x^2 – x – 3 = 0?

Sep 9, 2015

The rational root theorem only helps us find that ${x}^{3} - {x}^{2} - x - 3 = 0$ has no rational roots.

The only Real root is:

$x = \frac{1}{3} + \frac{\sqrt[3]{46 + 6 \sqrt{57}}}{3} + \frac{\sqrt[3]{46 - 6 \sqrt{57}}}{3} \approx 2.1304$

#### Explanation:

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

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $q \ne 0$, $p$ is a divisor of the constant term $- 3$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots of $f \left(x\right) = 0$ are:
$\pm 1$, $\pm 3$

Let's try them:

$f \left(1\right) = 1 - 1 - 1 - 3 = - 4$
$f \left(- 1\right) = - 1 - 1 + 1 - 3 = - 4$
$f \left(3\right) = 27 - 9 - 3 - 3 = 12$
$f \left(- 3\right) = - 27 - 9 + 3 - 3 = - 36$

So $f \left(x\right) = 0$ has no rational roots and the rational root theorem cannot help us find the roots that do exist.

Using a simple Tschirnhaus transformation ($t = x - \frac{1}{3}$) and Cardano's method, I found:

$x = \frac{1}{3} + \frac{\sqrt[3]{46 + 6 \sqrt{57}}}{3} + \frac{\sqrt[3]{46 - 6 \sqrt{57}}}{3} \approx 2.1304$