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

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Answer 1 · 2017-05-03T10:33:30

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

$x^{3} - 3 x^{2} - 4 x + 12 \leftarrow$ $x^3 -3x^2 -4x +12" "larr$ group into pairs

$= (x^{3} - 3 x^{2}) + (- 4 x + 12) \leftarrow$ $=(x^3 -3x^2)+ (-4x +12)" "larr$ there must be a $+$ $+$ between.

$x =^{2} (x - 3) - 4 (x - 3) \leftarrow \div - 4$ $x=^2(x-3)color(blue)(-4(x-3))" "larr div color(blue)(-4)$ out as the factor to change the signs.

$= (x - 3) (x^{2} - 4) \leftarrow$ $=(x-3)(x^2-4)" "larr$ factors further as difference of squares.

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

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