# How do you factor x^3-3x^2+4x-12=0?

$\left({x}^{2} + 4\right) \left(x - 3\right) = 0$
${x}^{3} - 3 {x}^{2} + 4 x - 12 = {x}^{2} \left(x - 3\right) + 4 \left(x - 3\right) = \left({x}^{2} + 4\right) \left(x - 3\right) = 0$[Ans]