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

${x}^{3} - 4 {x}^{2} - 2 x + 8 = {x}^{2} \left(x - 4\right) - 2 \left(x - 4\right) = \left(x - 4\right) \left({x}^{2} - 2\right)$
$= \left(x - 4\right) \left(x + \sqrt{2}\right) \left(x - \sqrt{2}\right)$