How do you use the rational roots theorem to find all possible zeros of x^3-3x-2?

Jun 2, 2016

${x}^{3} - 3 x - 2$ has zeros $x = - 1$ with multiplicity $2$ and $x = 2$.

Explanation:

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

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 $- 2$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$

We find:

$f \left(- 1\right) = - 1 + 3 - 2 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor of $f \left(x\right)$:

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

$- 1$ is also a zero of ${x}^{2} - x - 2$:

If $x = - 1$ then ${x}^{2} - x - 2 = 1 + 1 - 2 = 0$

Then we find:

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

So $x = 2$ is the last zero.