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I've found (maybe, maybe not, but it's not on this Wikipedia or this Wikipedia) that there is a subset of easily solved quartic polynomials of the form

$$0=x^4+2px^3+(p^2+p+2q)x^2+(2pq+p^2-1)x^1+(q^2+pq+q)x^0\tag1$$

You can break it down into the following form:

$$0=f(f(x))-x\tag2$$

where $f(x)=x^2+px+q$ and has 2 solutions easily found:

$$0=f(x)-x\tag3$$

$$x=\frac{-(p-1)\pm\sqrt{(p-1)^2-4q}}{2}\tag4$$

And from here, the second 2 solutions to the original quartic polynomial are easily found through polynomial division:

$$0=\frac{x^4+2px^3+(p^2+p+2q)x^2+(2bc+b^2-1)x^1+(q^2+pq+q)x^0}{(x-r_0)(x-r_1)}\tag0$$

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 $(3)$,

$$0=f(x)-x$$

$$x=f(x)\tag5$$

Apply the operation $f$ to both sides

$$f(x)=f(f(x))\tag6$$

The left side of $(6)$ is equivalent to the right side of $(5)$. This gives us

$$x=f(f(x))\tag2$$

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.

$$0=x^4+ax^3+bx^2+cx+d$$

Equating parts gives us:

$$2p=a$$

$$p^2+p+2q=b$$

$$2pq+p^2-1=c$$

$$q^2+pq+q=d$$

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:

$$\frac{x^4+2px^3+(p^2+p+2q)x^2+(2bc+b^2-1)x^1+(q^2+pq+q)x^0}{x^2+(p-1)x+q}$$

$$=x^2+(p+1)x+\frac{p+1}{x^2+(p-1)x+q}$$

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:

$$x=t+h$$

Place this into the general quartic polynomial (make it a polynomial of $t$) to get

$$0=(t+h)^4+a(t+h)^3+b(t+h)^2+c(t+h)+d$$

Attempting to expand it and such:

$$0=t^4+(4h+a)t^3+(6h^2+3ah+b)t^2+(4h^3+3ah^2+2bh+c)t+h^4+ah^3+bh^2+ch+d$$

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.

EDIT

I have managed to perform the division, and these are my results:

$$0=\frac{x^4+2px^3+(p^2+p+2q)x^2+(2bc+b^2-1)x^1+(q^2+pq+q)x^0}{x^2+(p-1)x+q}$$

$$=x^2+\left(1+2p-q+\frac{q^2}{p+q}\right)x+q+p+1$$

And so, the $4$ solutions to my special quartic are:

$$x_{1,2}=\frac{1-p\pm\sqrt{(p-1)^2-4q}}2$$

$$x_{3,4}=\frac{q-\frac{q^2}{q+p}-2p-1\pm\sqrt{(q-\frac{q^2}{q+p}-2p-1)^2-4(q+p+1)}}{2}$$

I went ahead and found the special solution when $q=-p$:

$$x_1=1$$

$$x_2=q$$

$$x_{3,4}=\frac{3q^2\pm\sqrt{9q^4-16q^2-16q-4}}{4q+2}$$

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 $q=-p=-\frac12$?

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I'm not sure that your question is, but I will make a remark on the 'generality' of your quartic:

$$x^4+2px^3+(p^2+p+2q)x^2+(2pq+p^2-1)x^1+(q^2+pq+q)=0$$

This contains only $2$ parameters, $p$ and $q$.

However, we can easily see that there is much more general $4$ parameter quartic equation with 'simple' solutions:

$$x^4+2ax^3+(a^2+2b+c)x^2+a(2b+c)x+(b^2+bc+d)=0 \tag{1}$$

The solutions are:

$$x_{1,2,3,4}=-\frac{1}{2} \left(a \pm \sqrt{a^2-4b-2(c \pm \sqrt{c^2-4d})} \right)$$


If we take:

$$a=c=p,~~~~b=d=q$$

We obtain:

$$x^4+2px^3+(p^2+p+2q)x^2+(2pq+p^2)x+(q^2+pq+q)=0$$

Which is very similar to your equation (except for one $-x$ term you have).

Actually, this $-x$ term makes everything much more complicated than it needs to be. I'm not sure how you go from $(2)$ to $(3)$ - it's not obvious to me at all.

Moreover, to check your solution, did you try numerical examples? I would check for you if I have the time, but it's an easy way to find if there is a mistake somewhere.


The method of obtaining $(1)$ is based on the same principle as yours, only we use different quadratic functions:

$$(x^2+ax+b)^2+c(x^2+ax+b)+d= \\ = x^4+2ax^3+(a^2+2b+c)x^2+a(2b+c)x+(b^2+bc+d)$$

We solve the 'outer' quadratic equation first, and then the 'inner' one to get the four solutions.


We can even make a $5$ parameter quartic by replacing $x \to x+f$ in the above for some constant $f$:

$$[(x+f)^2+a(x+f)+b]^2+c[(x+f)^2+a(x+f)+b]+d=0$$

Now the four solutions will be:

$$x_{1,2,3,4}=-f-\frac{1}{2} \left(a \pm \sqrt{a^2-4b-2(c \pm \sqrt{c^2-4d})} \right)$$


Edit

Actually, the correct 'user friendly' way would be to define the solution to $(1)$ in terms of coeffitients of the quartic and the condition that it will work. It goes like this:

$$x^4+Ax^3+Bx^2+Cx+D=0 \tag{2}$$

The only condition we need for our solution to be valid is this:

$$B=\frac{A^2}{4}+\frac{2C}{A} \tag{3}$$

Then the solution will be:

$$x_{1,2,3,4}=-\frac{1}{4} \left(A \pm \sqrt{A^2-\frac{16C}{A} \pm 16\sqrt{\frac{C^2}{A^2}-D}} \right) \tag{4}$$

So note that we only have 3 free parameters here, and the form of the equation is determined by the condition $(3)$.

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  • $\begingroup$ Thanks for the method. I actually read about something like this while sifting through the decomposition Wikipedia page. $\endgroup$ – Simply Beautiful Art Jun 19 '16 at 0:45
  • $\begingroup$ @SimpleArt, see my edit please for the clarification about 'free parameters' $\endgroup$ – Yuriy S Jun 19 '16 at 1:18
  • $\begingroup$ Well, I guess that's cool and all. To be honest, I've already figured out how to solve the general quartic by making both sides perfect squares or something like that. But nice answer. $\endgroup$ – Simply Beautiful Art Jun 19 '16 at 14:57
  • $\begingroup$ @SimpleArt, I look forward to seeing your results then, if you deside to post them $\endgroup$ – Yuriy S Jun 20 '16 at 6:41
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    $\begingroup$ @SimpleArt, agreed. Still, a very nice and clear method $\endgroup$ – Yuriy S Jun 28 '16 at 11:48
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The general solution to a quartic equation involves messy combinations of square and cube roots. If you try to combine all the steps in the general method into one formula like the quadratic formula, that formula will look like a Rube Goldberg machine (e.g., http://www.digitaltrends.com/cool-tech/best-rube-goldberg-machines/).

If you follow the general method (http://mathworld.wolfram.com/QuarticEquation.html) for a quartic equation with integer coefficients, you come to a resolvent cubic equation (https://en.wikipedia.org/wiki/Resolvent_cubic) that can also be expressed with integer coefficients. You should seek a rational root of this equation. If you find one, then the cube root radicals are gone and the quartic equation's roots are much more tractable. This is also the condition an irreducible quartic equation must satify if the roots are to be constructible with an unmarked straightedge and compasses.

Let's look at an example most SE users know well. The equation is $x^4+x^3+x^2+x+1=0$. Work out the resolvent cubic, and in terms of the resolvent cubic variable $u$ we have $4096u^3+1280u^2-720u-25=0$. This has a rational root $u=5/16$, so the full quartic roots can be rendered without cube roots and, of course, the roots of this fifth-order cyclotomic polynomial can be constructed in the complex plane.

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