# How do you find the zeroes of f(x)=x^3+10x^2-13x-22 ?

Aug 5, 2015

Use the rational roots theorem to find possible roots, evaluate $f \left(x\right)$ for those candidates to find roots $x = - 1$ and $x = 2$. Deduce the third root $x = - 11$.

#### Explanation:

By the rational roots theorem, any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ where $p$, $q$ are integers, $q \ne 0$, $p$ a divisor of the constant term $22$ and $q$ a divisor of the coefficient $1$ of the term of highest degree ${x}^{3}$.

So the only possible rational roots are:

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

$f \left(1\right) = 1 + 10 - 13 - 22 = - 24$
$f \left(- 1\right) = - 1 + 10 + 13 - 22 = 0$
$f \left(2\right) = 8 + 40 - 26 - 22 = 0$

So far, that gives $\left(x + 1\right)$ and $\left(x - 2\right)$ as factors of $f \left(x\right)$.

So the remaining factor must be $\left(x + 11\right)$ to get the correct coefficient for ${x}^{3}$ and the constant term $- 22$.

$f \left(x\right) = \left(x + 1\right) \left(x - 2\right) \left(x + 11\right)$

Let us check:

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

So the roots of $f \left(x\right) = 0$ are $x = - 1$, $x = 2$ and $x = - 11$