Precalculus
posted by Arielt .
Suppose that a polynomial function of degree 4 with rational coefficients has i and (3 + square root of 3)as zeros find the other zeros

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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