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Let $f(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_0$ be a polynomial with integer coefficients and irreducible over $\mathbb{Q}$. For $n \geq 3$, $f$ is generically unsolvable by radicals. Indeed, most irreducible polynomials of degree $2n$ have Galois group $S_{2n}$, which is unsolvable for $n \geq 3$. I am looking for a characterization, possibly in terms of the roots of $f$, for when $f$ is solvable by radicals.

For the $n = 3$ case, a very satisfactory answer is given in this paper: http://www.sciencedirect.com/science/article/pii/S002186930098428X

In particular, an answer such as "the Galois group of $f$ is solvable if and only if one of a small list of Galois resolvents has a rational root" would be very satisfactory.

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    $\begingroup$ I think it is unlikely that the Galois resolvent method would give a 'small list' for any $n$, due to the plethora of solvable subgroups of $S_{2n}$ for large $n$.. $\endgroup$ Commented Nov 20, 2015 at 0:53

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There is certainly no satisfactory answer to the question, one reason being that $S_{2n}$ has many solvable transitive subgroups, as pointed out by Walter Neff in the comments.

Another reason is that if there were a manageable criterion, then we could handle polynomials $g(x)$ of odd degree as well by considering $f(x)=g(x^2-a)$ with $a\in\mathbb Q$ such that $f(x)$ is irreducible. (The existence of such an element $a$ isn't hard to establish.)

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