Given $\alpha,\,\beta,\,\gamma$ three roots of $g(x)\in\mathbb Q[x]$, a monic polynomial of degree $3$. We know that $\alpha+\beta+\gamma=0$, $\alpha^2+\beta^2+\gamma^2=2009$ and $\alpha\,\beta\,\gamma=456$. Is it possible to find the polynomial $g(x)$ only from these?
I've been working with the degree of the extension $\mathbb Q \subseteq \mathbb Q(\alpha,\,\beta,\,\gamma)$. I've found that it must be $3$ because $g(x)$ is the irreducible polynomial of $\gamma$ over $\mathbb Q(\alpha,\,\beta)$. But there is something that doesn't hold, there must be some of these roots that are not algebraic or something. May be this approach is totally wrong. Is there anyone who knows how to deal with this problem?
Answer
The polynomial is $(x-a)(x-b)(x-c)$ with the roots being $a,b,c$. By saying "three roots" you imply all these are different. Note that when multiplied out and coefficients are collected you have three symmetric functions in the roots. For example the constant term is $-abc$, while the degree 2 coefficient is $-(a+b+c)$. The degree 1 coefficient is $ab+ac+bc$, which can be written as
$$\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}.$$
So it looks like you can get all the coefficients of the monic from the givens you have.
Note: Just saw copper.hat's remark, essentially saying what's in this answer. I'll leave it up for now in case the poser of the question needs it (or can even use it...)
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