How do you factor $y = x^{3} - 3 x^{2} - 4 x + 12$ $y= x^3-3x^2-4x+12$ ?

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Answer 1 · 2018-05-17T22:27:52

$y = (x + 2) (x - 2) (x - 3)$ $y=(x+2)(x-2)(x-3)$

$y = x^{3} - 3 x^{2} - 4 x + 12$ $y=x^3-3x^2-4x+12$

$y = x^{2} \cdot (x - 3) - 4 \cdot (x - 3)$ $y=x^2*(x-3)-4*(x-3)$

$y = (x - 3) \cdot (x^{2} - 4)$ $y=(x-3)*(x^2-4)$

$y = (x + 2) (x - 2) (x - 3)$ $y=(x+2)(x-2)(x-3)$

How do you factor y= x^3-3x^2-4x+12y=x3−3x2−4x+12y= x^3-3x^2-4x+12 ?