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

The rational root theorem states that any rational root of a polynomial will be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
In our case $q = 1$ and $p = - 3$ hence possible roots are

$1 , - 1 , - 3 , 3$

But if you check none of the values above are roots for
${x}^{3} - {x}^{2} - x - 3 = 0$.
So the rational root theorem cannot help us here.

Dec 26, 2015

You can't. You can only use the rational root theorem to show that it has no rational roots.

#### Explanation:

By the rational root theorem, any rational roots of ${x}^{3} - {x}^{2} - x - 3 = 0$ must be expressible in the form $\frac{p}{q}$ for integers $p$ and $q$ with no common factor apart from $\pm 1$, where $p$ is a factor of the constant term $- 3$ and $q$ a factor of the coefficient $1$ of the leading term.

That means that the only possible rational factors are:

$\pm 1$, $\pm 3$

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

We find:

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

So ${x}^{3} - {x}^{2} - x - 3 = 0$ has no rational roots.

In fact it has one Real root:

$x = \frac{1}{3} \left(1 + {\left(46 - 6 \sqrt{57}\right)}^{\frac{1}{3}} + {\left(46 + 6 \sqrt{57}\right)}^{\frac{1}{3}}\right)$

and two Complex roots:

$x = \frac{1}{3} \left(1 + \omega {\left(46 - 6 \sqrt{57}\right)}^{\frac{1}{3}} + {\omega}^{2} {\left(46 + 6 \sqrt{57}\right)}^{\frac{1}{3}}\right)$

$x = \frac{1}{3} \left(1 + {\omega}^{2} {\left(46 - 6 \sqrt{57}\right)}^{\frac{1}{3}} + \omega {\left(46 + 6 \sqrt{57}\right)}^{\frac{1}{3}}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.

These can be found using Cardano's method or similar.