SOLUTION: x^3-3x^2-4x+12=0 Please explain how to solve

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Question 505491: x^3-3x^2-4x+12=0 Please explain how to solve
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x^3-3x^2-4x+12=0

(x^3-3x^2)+(-4x+12)=0

x^2(x-3)+(-4x+12)=0

x^2(x-3)-4(x-3)=0

(x^2-4)(x-3)=0

(x-2)(x+2)(x-3)=0

x-2 = 0, x+2 = 0, or x-3 = 0

x = 2, x = -2, or x = 3