Solvable by radicals

Remarks

Equivalently, every root of $f$ lies in some radical extension of $K$, since the compositum of finitely many radical extensions inside a common field is radical. Example: $f = x^3 - 2$ over $\Q$ splits in $R = \Q(\omega, \sqrt[3]{2})$, with $\omega = e^{2\pi i/3}$. This $R$ is radical, since $\omega^3 = 1 \in \Q$ and $(\sqrt[3]{2})^3 = 2 \in \Q(\omega)$, so $f$ is solvable by radicals. Non-example: $f = x^5 - 4x + 2$ over $\Q$ is not solvable by radicals, because its Galois group is $S_5$, which is not solvable; see the quintic equation. In characteristic $0$, Galois' theorem on solvability says that $f$ is solvable by radicals if and only if its Galois group is solvable.