# How do you find all the zeros of x^3-4x^2-44x+96?

Aug 14, 2016

This cubic has zeros: $2$, $8$ and $- 6$

#### Explanation:

$f \left(x\right) = {x}^{3} - 4 {x}^{2} - 44 x + 96$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $96$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 8 , \pm 12 , \pm 16 , \pm 24 , \pm 32 , \pm 48 , \pm 96$

Trying each in turn, we find:

$f \left(2\right) = 8 - 16 - 88 + 96 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{3} - 4 {x}^{2} - 44 x + 96 = \left(x - 2\right) \left({x}^{2} - 2 x - 48\right)$

To factor the remaining quadratic find a pair of factors of $48$ which differ by $2$. The pair $8 , 6$ works. Hence:

${x}^{2} - 2 x - 48 = \left(x - 8\right) \left(x + 6\right)$

Hence zeros: $8$ and $- 6$