Reduce quintic equation

Question

If we have the general quintic equation $$ax^5+bx^4+cx^3+dx^2+ex+f=0$$ we can vanish the quartic term by doing the substitution $x=y-b/5a$. The question I wanna ask is if there is a possible way to do a substitution so the quarticc and the quadratic term is vanished so that we are left with the equation $$ay^5+by^3+cy+d=0$$ ($a$,$b$,$c$,$d$ not the same of course)
Thank you for your time.
If someone can fix the latex it would be great, thank you.

Answer 1

Let us see directly why this is impossible. I divide through by $a$ (or equivalently, consider $a = 1$) for ease of notation.

We start with $x^5 + bx^4 + cx^3 + dx^2 + ex + f$, and we will be doing a substitution $x \mapsto x - \lambda$. Let's see what constraints we get from wanting to remove the quartic term. So we collect the coefficients of $x^4$.

The $x^5$ term gives us $-5\lambda x^4$, and the $bx^4$ term gives us $bx^4$ (i.e. it doesn't change). We want these to cancel, so we want $bx^4 - 5\lambda x^4 = 0$, or for $\lambda = \frac{b}{5}$. This is completely forced, and no other substitution will remove the quartic term in general.

So doing any other substitution will not remove the quartic term, and this substitution does not in general remove the quadratic term (which I do not show, because it's an annoying computation - alternatively, choose just about any quintic and it will give a counterexample).

So the answer is no, it is not possible to do a generic substitution such that both the quartic and quadratic terms vanish.