# How do you find the zeroes of p(x)= x^3-x^2-10x-8?

Aug 11, 2018

$p \left(x\right)$ has zeros $- 1$, $4$ and $- 2$

#### Explanation:

Given:

$p \left(x\right) = {x}^{3} - {x}^{2} - 10 x - 8$

By the ratonal zeros theorem, any rational zeros of $p \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 8$ and $q$ a divisor of the coefficient $1$ f the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 8$

Trying each in turn, we find:

$p \left(\textcolor{b l u e}{- 1}\right) = {\left(\textcolor{b l u e}{- 1}\right)}^{3} - {\left(\textcolor{b l u e}{- 1}\right)}^{2} - 10 \left(\textcolor{b l u e}{- 1}\right) - 8$

$\textcolor{w h i t e}{p \left(- 1\right)} = - 1 - 1 + 10 - 8 = 0$

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

${x}^{3} - {x}^{2} - 10 x - 8 = \left(x + 1\right) \left({x}^{2} - 2 x - 8\right)$

We can factor the remaining quadratic by finding a pair of factors of $8$ which differ by $2$, namely $4$ and $2$, and hence:

${x}^{2} - 2 x - 8 = \left(x - 4\right) \left(x + 2\right)$

So the other two zeros are $x = 4$ and $x = - 2$.