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

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Answer 1 · 2015-05-23T11:54:06

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

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

By synthetic division we obtain
$p(x) = (x-1)(x^3-3x^2-4x+12)$

Focusing on $(x^3-3x^2-4x+12)$
we notice the grouping
$x^2(x-3) -4(x-3)$

$=(x^2-4)(x-3)$

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

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

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

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