# Exactly what does it mean that the quintic unsolvable

I am trying to get my head around Galois theory and the unsolvability of the general quintic (or equations of higher degree). The fundamental theorem of algebra states that a polynomial equation of degree n has exactly n solutions. So the quintic has 5 solutions, this would mean $x^{5} + a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0} = (x-R)(x^{4} + b_{3}x^{3} + b_{2}x^{2} + b_{1}x + b_{0})$ for some complex number $R$. And then a degree 4 polynomial equation is solvable. But the quintic is not solvable. It feels like I am missing something here.

Sometimes you read formulations like there is no radicals in the coefficients of the equation that solves the equation. But what about radicals with other numbers than the coefficients?

• – Crostul Jul 2 '16 at 12:18
• "Solvable" in this conetext means : "Solveable by radicals". You must determine the solutions only by taking roots and applying addition, subtraction, multiplication and division. – Peter Jul 2 '16 at 12:24
• Usually, only polynomials with integer coefficients are considered. Then, the so-called galois-group over $\mathbb Q$ of this polynomial (usually irreducible) is determined. Then, the polynomial is solvable by radicals if and only if the galois-group is solvable. – Peter Jul 2 '16 at 12:26
• The coefficients are called 'parameters', and it basically mean that they are the reason the roots are what they are. Thus, it only makes sense to have solutions in terms of the coefficients. – Simply Beautiful Art Jul 2 '16 at 12:27
• You could think about it like this: If you could factor some $(x-R)$ out, then the quintic is solvable. But you can't know what $R$ exactly is, since you can't solve for it in terms of radicals. It is like the difference between $\pi r^2$ and $3.14r^2$ – Simply Beautiful Art Jul 2 '16 at 12:29

The Abell-Ruffini theorem does not say that an equation of degree $>4$ is unsolvable, but that , for such equations , there does not exists in general a formula that gives the solutions in term of rational operations and radicals.
• The Abel-Ruffini theorem is stronger than the claim that there is no general fomula for polynomials with degree $5$ or higher. This claim would leave the possibility that every equation would allow some special solving method. – Peter Jul 2 '16 at 12:35
• In fact, a random equation of degree $5$ or higher is very likely to be non-solvable by radicals. – Peter Jul 2 '16 at 12:37