SOLUTION: Solve the polynomial: x^4+x^3-6x^2-14x-12=0

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Question 1010528: Solve the polynomial:
x^4+x^3-6x^2-14x-12=0
Found 2 solutions by stanbon, Boreal:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the polynomial:
x^4+x^3-6x^2-14x-12=0
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graph%28400%2C400%2C-10%2C10%2C-50%2C50%2Cx%5E4%2Bx%5E3-6x%5E2-14x-12%29
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You see Real solutions at x = -2 and x = 3
-2)....1....1....-6....-14....-12
.......1...-1....-4....-6....|..0
3).....1....2.....2....|..0
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Solve x^2 + 2x + 2 for the complex roots::
x = [-2 +- sqrt(4-2)]/2
x = -1 + (1/2)sqrt(2) or x = -1 - (1/2)sqrt(2)
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Cheers,
Stan H.
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Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x^4+x^3-6x^2-14x-12=0. Roots possible include +/- 12,6,4,3,and 2. It has one sign change, so it has one positive root. f(-x) +/-/-/+/-, so there are 3 or 1 negative roots.
1 ;;1;;-6;;-14;;-12
1;;-1;;-4;;-6 synthetic division using -2 as a root or (x+2)
now we have x^3-x^2-4x-6
1;;-1;;-4;;-6 try 3
1;;2;;2;;0. That works (x-3)
x^2+2x+2. That has complex roots.
graph%28300%2C200%2C-10%2C10%2C-10%2C100%2Cx%5E4%2Bx%5E3-6x%5E2-14x-12%29