# How do you find the zeroes of p(x)= x^4-4x^3-x^2+16x-12?

##### 1 Answer
May 23, 2015

By observation we can note that
if $p \left(x\right) = {x}^{4} - 4 {x}^{3} - {x}^{2} + 16 x - 12$
then
$p \left(1\right) = 0$

Therefore $\left(x - 1\right)$ is a factor of $p \left(x\right)$

By synthetic division we obtain
$p \left(x\right) = \left(x - 1\right) \left({x}^{3} - 3 {x}^{2} - 4 x + 12\right)$

Focusing on $\left({x}^{3} - 3 {x}^{2} - 4 x + 12\right)$
we notice the grouping
${x}^{2} \left(x - 3\right) - 4 \left(x - 3\right)$

$= \left({x}^{2} - 4\right) \left(x - 3\right)$

and using the difference of squares
$= \left(x + 2\right) \left(x - 2\right) \left(x - 3\right)$

Therefore a complete factoring of $p \left(x\right)$
$= \left(x - 1\right) \left(x + 2\right) \left(x - 2\right) \left(x - 3\right)$

and the zeroes of $p \left(x\right)$ are
$\left\{1 , - 2 , 2 , 3\right\}$