Posted by Arielt on Sunday, March 14, 2010 at 1:25pm.
There must be four roots total and complex numbers have conjugates that are roots. One other root is therefore
-i .
If -3 + sqrt3, is a root, the another root is -3 - sqrt3, as a consequence of the +/-sqrt(b^2 - 4ac) in the quadratic equation.
The polynomial must be a multiple of
(x^2 +1)(x +3 -sqrt3)(x +3 +sqrt3) = 0
(x^2+1)[(x+3)^2 -3] = 0
(x^2+1)(x^2 +6x +6) = 0
x^4 +6x^3 +7x^2 +6 = 0
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