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.
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 $f$.