SOLUTION: Please help me solve the equation. X'3 + 3x'2 - 4x - 12 =0

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Question 964817: Please help me solve the equation.
X'3 + 3x'2 - 4x - 12 =0

Found 2 solutions by Alan3354, josgarithmetic:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve the equation.
X'3 + 3x'2 - 4x - 12 =0
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Use ^ (Shift 6) for exponents.
X^3 + 3x^2 - 4x - 12 = 0
Factor by grouping.
Notice that the coeff's of the 2nd 2 terms are -4 times the 1st 2.
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X^3 + 3x^2 - 4x - 12 = x^2*(x-3) - 4*(x-3)
--> (x^2-4)*(x+3) = 0
--
x+3 = 0
x = -3
=================
x^2-4 = 0
x = -2,+2

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
x^3 + 3x^2 - 4x - 12 =0
x%5E3+%2B+3x%5E2+-+4x+-+12+=0

x%5E3-4x%2B3x%5E2-12=0
x%28x%5E2-4%29%2B3%28x%5E2-4%29=0
%28x%2B3%29%28x%5E2-4%29=0
All through factoring. Easier now.