I've got to proof the following:(adsbygoogle = window.adsbygoogle || []).push({});

Let f be a monic polynomial in Q[X] with deg(f) = n different complex zeroes. Show that the sign of the discriminant of f is equal to (-1)^s, with 2s the number of non real zeroes of f.

I know the statement makes sense, because the discriminant is a symmetric polynomial over Q, so it can be written as a polynomial in elementary symmetric polynomials.

The question seems to suggest that the complex zeroes always come in pairs of a zero and its conjugate. But even if this is true, i still don't know how to proceed :(

Can anyone give me a hint?

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# Homework Help: Discriminant is a symmetric polynomial

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