I know that fifth and higher order polynomial equations cannot be solved in terms of radicals in general. My question is about two special cases and I’m not sure where to look. I’m interested in solutions to

ax^n + bx + c = 0 (n > 4)

ax^n + bx^2 + c (n > 4 and odd)

That is equations with only the nth, first, and zeroth power terms or
only the the nth, second, and zeroth power terms.

(And yes I know
if a, b, or c = 0 or they have easy solutions
if n is even in the second case, it’s the same as the first case for x^2)

Can either of these be solved in general (for real coefficients if it matters).

This isn’t homework which for me ended some 30+ years ago.

If by “can be solved” you mean “can be solved in radicals”, then for the first, the answer is no. x[sup]5[/sup] - x - 1 is an example of a polynomial whose roots don’t lie in the closure of the rationals by radicals.

Actually I’m interested in can the answer be expressed in “closed form” that is using commonly defined functions. I suspect this means the same thing, but am not sure. I realize that “commonly defined functions” is a rather loose term so I didn’t use it at first as I suspected the answers are the same in either case.

For the purposes that you’re asking, x^5 - x - 1 can’t be solved. One can define new functions expressly for the purpose of solving problems like that, but then, one can do that for any problem at all.

Incidentally, I said this only answered your first question, but it answers your second question as well, by your own observation: replacing x by x[sup]2[/sup], we also find that x[sup]10[/sup] - x[sup]2[/sup] - 1 has no roots in the radical closure of the field of rationals.

That’s not quite enough to rule out any “closed form” expression in terms of “commonly defined functions”, in itself, but, while I don’t know any proof offhand, I am confident that even adding the natural exponential and logarithm (and, via those, the trigonometric functions and their inverses) is not enough to get the roots of this polynomial.

It doesn’t answer my second question, quite, since I specified n odd in that case specifically because I realized that ax[sup]2n[/sup] + bx[sup]2[/sup]+c was equivalent in x[sup]2[/sup] to the first case. But I’m guessing those can’t be solved either.

Probably what you want are the so-called elementary functions; these are just what I referred to before, what you can build out of the field operations plus the natural exponential and logarithm. But they include pretty much any function I’ve ever seen considered to be in closed-form in, say, a college calculus class.

Eh, depending on the application, one might also consider the Bessel functions, or the Beta and Gamma functions, or the elliptic functions, or a few other usual suspects, to be “closed-form”. As in, once you’ve gotten the answer down to one of those, you can be sure that you’re not simplifying it any further to get rid of those special functions, so you might as well be happy with what you have.

Sure. As you said, and as I’ve often argumentatively stressed myself, depending on the application, you may find call to take different things to count as closed-form, anything at all being closed-form from some point of view.

But, if one wants to draw a precise circle around what the OP probably has already in mind, then I think the elementary functions capture well the demarcation between what you’d see presented as a closed-form expression in a typical undergraduate basic calculus class and what you wouldn’t. I doubt the OP has Bessel functions in mind, for example, any more than they have Bring radicals in mind.

If you’re interested in solutions by radicals, then it’s necessary and sufficient that Gal(K/Q) be solvable, where K is the splitting field of your rational polynomial f(X). (The proof is essentially just noting that a “radical” extension L’/L has Gal(L’/L) cyclic, and K is a tower of such extensions.) Since S_n isn’t solvable for n > 4, a general nth degree polynomial for n > 4 isn’t solvable by radicals, but some individual polynomials are.

If you’re talking about adding in a set of special functions instead of just extensions by radicals, then you might be able to do a similar analysis by considering your polynomials over some new base field K’ and classifying all Galois extensions L’/K’ that meet whatever the new criteria replacing solvability is. Differential Galois theory sounds closer to what you’re looking for, but I don’t know much about the subject.

Actually Bessle functions would be fine, though I doubt they’d help here at all. I’d take, say any function in Abromowitz and Stegum (sp?) though I can’t imagine any would help with polynomials. I think of all those as helping with differential equations.