So, we know there are generally no solutions to an arbitrary quintic polynomial (EDIT from comments: Of course, I mean you can't write the roots using radicals, not that the roots don't exist). I don't know Galois theory, so I understand this result by thinking about permutations of roots and coefficients (this video is awesome!). To me it seems to boil down to the fact that when you write a polynomial in terms of coefficients and roots:
$$x^5 + ax^4 + bx^3+cx^2+dx+e=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)$$
There is a natural ordering of coefficients, but not a natural ordering of the roots (you can permute the $x_i$'s and get the same set of coefficients, but if you switch the coefficients, you'll change the roots). That's from the basic property of commutativity of multiplication.
So...what if you considered a "noncommutative polynomial", in which
$$(x-x_1)(x-x_2)\neq (x-x_2)(x-x_1)$$
Since this would imply a natural order to the roots (and switching the roots would result in a different set of coefficients), does that mean these higher order polynomials would have solutions? Is there any work in this area?
I can see some simple and immediate implications of these things - for example, you need more coefficients, since polynomials can be written like
$$x^2+ax+xb+c=0$$
(even worse, I guess it's possible you have to write $x^2=dx^2+xex+x^2f$ or something.)
Ok, I just thought of the answer...my intuition was right, it's far easier to find the roots of a "noncommutative polynomial" (although I still don't know how generally interesting this concept is). Just think about writing it out for the quadratic:
$$(x-x_1)(x-x_2)=x^2-x x_2 - x_1 x + x_1 x_2=0$$
Now compare this to the most general form of a quadratic:
$$Ax^2 + xBx + x^2 C + Dx + xE + F =0$$
And we can read off the roots from the coefficient list!
$$A=1,~E=-x_2,~D=-x_1,~F=x_1x_2$$
Of course, there's some ambiguity here, since $C=1$ is also fine, and when I write "1" I mean "the noncommutative unit" I guess.
I would still like to hear if anyone finds anything else kind of interesting about this, but this is basically the answer I was looking for - I just didn't think too carefully about it before posting!