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

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Answer 1 · 2015-05-23T11:54:06

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

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

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

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

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

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

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

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

How do you find the zeroes of p(x)=x4−4x3−x2+16x−12p(x)= x^4-4x^3-x^2+16x-12?