How do you factor completely $x^3-4x^2-2x+8$ ?

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Answer 1 · 2015-04-12T15:28:32

The passages to do are:

$x^3-4x^2-2x+8=x^2(x-4)-2(x-4)=(x-4)(x^2-2)$

and... if you want:

$=(x-4)(x+sqrt2)(x-sqrt2)$

How do you factor completely x^3-4x^2-2x+8x^3-4x^2-2x+8?