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December 21, 2014

December 21, 2014

Posted by **Arielt** on Sunday, March 14, 2010 at 1:25pm.

- Precalculus -
**drwls**, Sunday, March 14, 2010 at 3:13pmThere 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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