Let $F$ be a field, I asked myself if $p(x)\in F[x]$ is solvable by radicals iff every irreducible factor is solvable by radicals.
My thoughts: If every irreducible factor is solvable by roots then it imply that there are field extensions that are solvable by roots, I would like to use the simple fact that the composition field of solvable extensions is also solvable, but in my case the extensions I have are not subfields of one field.
I don't know about the other direction, and I have a feeling it is not true but due to lack of examples that I know I can't think of a counter example.
Is this statement correct ? If so, how can we prove it ?