# What does insolvability of the quintic mean exactly?

Suppose I had the quintic equation $(x+1)(x+2)(x+3)(x+4)(x+5)=0$. Does the insolvability theory mean that I can only get approximations because the root is an generally an irrational number, or does it mean that even in this case I can only get an approximation to say -5 as well even though it is whole number and is an exact root?

I've read that Galois theory can tell you if a quintic polynomial is the type that can be solved exactly. My question is: Are these types of quintics simply the ones with integer roots? I would suspect that just because a quintic has integer coefficients doesn't mean they have integer roots.

Finally, If galois theory says that a particular quintic is of the exactly solvable type, what method is used to solve them exactly, besides the rational root theorem? I'm pretty sure it would be something like Cardano's method, adapted to a fifth degree equation. Where can you find this method?

-
The insolvability of the quintic is about quintic equations in general, not about particular quintics such as this one. It says that there does not exist an analogue of the quadratic formula for quintics (and note that the quadratic formula works for all quadratics). – Qiaochu Yuan Mar 5 '12 at 21:30
So would the methods to get answers for a quintic equation give you an exact answer if the answer is rational, or an apparently approximate answer if the answer is an irrational number that cannot be expressed using radicals? Is that what it is, or would those methods give you r = -4.9999999999999957 when we know the answer is -5 ? – Kenny Mar 5 '12 at 22:43
There is a general algorithm to find all rational roots of a polynomial. See en.wikipedia.org/wiki/Rational_root_theorem – Tib Mar 5 '12 at 22:51
You don't have a quintic equation. An equation has an equals sign, and an expression on each side of that equals sign. The equation $(x+1)(x+2)(x+3)(x+4)(x+5)=0$ can, of course, be solved exactly. The solutions to the equation $(x+1)(x+2)(x+3)(x+4)(x+5)=7$ cannot be expressed in terms of the 4 arithmetical operations, square roots, cube roots, fifth roots, etc. That's what's meant by insolvability (in this context); it means inexpressibility in terms of arithmetical operations on rationals and radicals. – Gerry Myerson Mar 5 '12 at 23:14
@Kenny: The unsolvability of the general quintic, as proved by Abel and probably earlier by Ruffini, is the result that there is no general formula for the roots in terms of the coefficients, where by formula one means something built up from the elementary operations of arithmetic, augmented by various $n$-th roots. Later, Galois exhibited individual specific quintics whose roots cannot be so expressed. That is a substantially harder result. – André Nicolas Mar 6 '12 at 5:50

## 1 Answer

Galois Theory assigns, to each polynomial, a mathematical structure called a group. A polynomial is solvable in radicals (that is, you can write down its roots in terms of its coefficients, the 4 arithmetical operations, and square roots, cube roots, etc.) if and only if the corresponding group is a "solvable" group. The definition of solvable group won't mean much to you if you haven't done a course in group theory; there should be a sequence of groups, starting with the trivial group and ending with the group corresponding to the polynomial, such that each group in the sequence is a "normal" subgroup of the next group, and the "quotient" of each group by the previous group is "commutative".

What's more, if the group corresponding to the polynomial is solvable, then this sequence of in-between groups, together with the quotient groups, can be used to construct the formula for the roots of the polynomial.

To expand this into something you could actually use to determine whether a polynomial is solvable in radicals, and to solve it if it is, takes a semester of advanced undergraduate level mathematics. Get yourself a good text on Galois Theory (assuming you have already done courses on Linear Algebra and an introductory Abstract Algebra course - if not, you'll have to study those first), read it, and enjoy.

-
I understand that the only way to tell if a polynomial is solvable when you don't know its roots is by using the galois theory. But can you know that it is solvable if you know it has only integer roots? – Kenny Mar 7 '12 at 12:15
So are you saying that galois theory can actually help you actually solve a solvable quintic in addition to telling you that it is solvable? – Kenny Mar 7 '12 at 12:46
Oh, you answered my last question in your second paragraph. Thanks. That's good news. – Kenny Mar 7 '12 at 12:54
If you know that a polynomial (of any degree) has only integer roots, then you can find those roots, and you don't need Galois Theory or anything very advanced. You can easily find an upper bound for the (absolute value) of the roots, and then test every integer up to that bound, or you can apply The Rational Root Theorem, q.v., to get a list of the possibilities that need checking. I know from your comments elsewhere that you don't like checking lots of cases, but this does show that it's solvable. – Gerry Myerson Mar 7 '12 at 21:45
I just want to know if you CAN find integer roots by the method of the second paragraph of your answer where if it is solvable (which you say it is), the sequence of in-between and quotient groups "can be used to construct the formula for the roots of the polynomial." I don't care if it is the easiest way or is even hopelessly complicated. I just want to know if it CAN be done this way. Can it be done that way? – Kenny Mar 11 '12 at 16:35