Finding if f(x)=3x6−5x5+x2f(x)=3x^6-5x^5+x^2 is solvable by radicals (in Q\mathbb{Q})

[MathNugget]

Reference for radical extensions: https://en.wikipedia.org/wiki/Radical_extension#:~:text=Solvability by radicals,-Radical extensions occur&text=In fact a solution in,a radical extension of K.

Short summary: a field extension K/k is simple radical if K=k[$\alpha$], and $\alpha^n=a$, for some $a \in k$ and an $n \in \mathbb{Q}$ (probably definition works over a bigger field, but let's make the problem easier).
A (general) radical extension is a set of simple radical extensions $F_1\subset F_2\subset F_3...$, with each $F_{i-1}\subset F_i$ simple radical...

Well, we first break the polynomial up a little: $f(x)=3x^6-5x^5+x^2=x^2(3x^4-5x^3+1)$. As $x^2$ provides nothing to the extension, I am left with a 4th degree polynomial. According to the internet, Abel–Ruffini theorem says "if the degree of the polynomial is less or equal to 4, it's definitely solvable by radicals". I guess this completes the proof? the field extension is now generated by some radicals...

 

[MaxWong]

Yes, this completes the proof.

 

[MathNugget]

Yes, this completes the proof.

Thank you.