# 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?

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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