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

1 Answer
EZ as pi
May 3, 2017

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

Explanation:

#x^3 -3x^2 -4x +12" "larr# group into pairs

#=(x^3 -3x^2)+ (-4x +12)" "larr# there must be a #+# between.

#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)" "larr# factors further as difference of squares.

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

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