Posted by Vic on Friday, February 18, 2011 at 8:38pm.
If there is a rational root, it would be +or- 1,2,3,4,6,8 or 12.
x = -2 works, so x+2 is a factor.
So, after one factoring step, and a bit of polynomial long division, you have
(x+2)(x^3 -x^2 -6x +18) = 0
Next, use the rational root theorm on the cubic factor. You will find that -3 is a root, making (x+3) a factor.
So now you have
x^4+x^3-8x^2+6x+36
(x+2)(x+3)(x^2 -4x +6)
The remaining two factors are complex conjugates.
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