You can break it down into the following form:
where $f(x)=x^2+px+q$ and has 2 solutions easily found:
And from here, the second 2 solutions to the original quartic polynomial are easily found through polynomial division:
After division, apply the quadratic formula to find easy solutions.
The uniqueness of this is that we take an $x$ out before decomposing, and when we decompose, we want it in the form $f(f(x))-x$.
To get from step $(2)$ to step $(3)$,
Apply the operation $f$ to both sides
The left side of $(6)$ is equivalent to the right side of $(5)$. This gives us
However, extra solutions may occur due to $f$ not being injective, but, solving the easier problem of $f(x)=x$ finds solutions that must be solutions to the original.
What I like about this solution is that you can easily determine if the given quartic is solvable through my method.
Equating parts gives us:
One can easily determine what $p$ could be from the first equalities and use the other three to verify if it is true.
This doesn't solve the general quartic of course (I think, read on) because there are only two unknowns $p,q$ which must relate to four unknowns $a,b,c,d$.
Now, I ran into a problem at $(0)$ where I couldn't easily simplify the quotient.
I started with $(x-r_0)(x-r_1)=x^2+(p-1)x+q$ and then proceeding with long division. And my result was this:
I am unsure if I made a mistake somewhere in the process of long division, so it would be nice if someone could try applying the division themselves to find if I made a mistake.
So by finding out what $p,q$ should be, we can determine if the quartic polynomial is solvable with my method. However, I thought of an idea that could possibly make my solution work for all quartics.
Consider the use of substitution:
Place this into the general quartic polynomial (make it a polynomial of $t$) to get
Attempting to expand it and such:
And I was wondering if there was some $h$ that allowed the general quartic to simplify down to my special case of the quartic polynomial.
So the two questions I pose concern if I did my division correctly and if there exists some $h$ that transforms the general quartic into my quartic and what that $h$ is.
And of course, feel free to point out if this is a well known or already known solution or any questions/comments you may have.
I have managed to perform the division, and these are my results:
And so, the $4$ solutions to my special quartic are:
I went ahead and found the special solution when $q=-p$:
And if you want the solution to the very very special case of $q=-p=-\frac12$, I pose it as my last question.
So my new questions are if there exists some $h$ that transforms the general quartic into mine and what is the special solution when $q=-p=-\frac12$?