Question 226626: Help me solve the following by factoring:
x^3-3x^2-4x+12=0
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! x^3 -3x^2 -4x +12 =0
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with 4 terms, normally try grouping
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x^2 ( x-3) -4(x-3)=0
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(x^2 -4) ( x-3) = 0,,,,,now factor difference of 2 squares
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(x+2)(x-2)(x-3)=0
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x= 2, -2, 3
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check,,,(2),,,(2)^3 -3(2)^2 -4(2) +12 =0
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8-12-8+12=0,,,,ok
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(-2),,,,(-2)^3 -3(-2)^2 -4(-2) +12 =0
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-8 -12 +8 +12 =0,,,,ok
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(3),,,,,,(3)^3 -3(3)^2 -4(3)+12 = 0
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27 -27 -12+12 = 0,,,,,ok
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