Suppose you have some polynomial $p(x)$ with rational coefficients in which at least one root is unsolvable by radicals, does this imply that all other roots of $p(x)$ are unsolvable by radicals?

I have been thinking about algebraic numbers for a while, and I found this question interesting. I tried a few examples, and I tried to think to a solution but all I concluded was this below.

If the constant factor is zero, and the polynomial has an unsolvable root, then I have an answer to the question since if the constant is zero, then it must have a root at zero.

This also implies that if the answer to my question is yes, then any polynomial with zero as the constant term, is solvable by radicals.

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    $\begingroup$ Clearly this question is interesting only for irreducible polynomials. $\endgroup$ – Crostul Jul 13 '14 at 18:10

The answer to your question is no for the trivial reason you already observed: suppose you have a polynomial $f(x)$ for which all roots are unsolvable, then $xf(x)$ has the solvable root $0$ but all others are unsolvable.

The situation changes if you assume that $f$ to be irreducible: then if one root can be expressed by radicals, the same is true for all of the roots. This is a direct consequence of the restatement of solvability in terms of the Galois group of $f$.


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