How do you use the rational roots theorem to find all possible zeros of f(x)=x^3+10x^2-13x-22 ?

Jun 5, 2016

$x = - 1$, $x = 2$ and $x = - 11$

Explanation:

$f \left(x\right) = {x}^{3} + 10 {x}^{2} - 13 x - 22$

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 $- 22$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 11$, $\pm 22$

Trying each in turn, we find:

$f \left(- 1\right) = - 1 + 10 + 13 - 22 = 0$

$f \left(2\right) = 8 + 40 - 26 - 22 = 0$

$f \left(- 11\right) = - 1331 + 1210 + 143 - 22 = 0$

So we have found all of the zeros:

$x = - 1$, $x = 2$ and $x = - 11$