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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?

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  • $\begingroup$ See en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem $\endgroup$ – Crostul Jul 2 '16 at 12:18
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    $\begingroup$ "Solvable" in this conetext means : "Solveable by radicals". You must determine the solutions only by taking roots and applying addition, subtraction, multiplication and division. $\endgroup$ – Peter Jul 2 '16 at 12:24
  • $\begingroup$ 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. $\endgroup$ – Peter Jul 2 '16 at 12:26
  • $\begingroup$ 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. $\endgroup$ – Simply Beautiful Art Jul 2 '16 at 12:27
  • $\begingroup$ 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 $3.14r^2$ $\endgroup$ – Simply Beautiful Art Jul 2 '16 at 12:29
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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.

In your example, if one solution of the quintic is know, we can factorize it as you say, but if this solution is not known, we cannot find the factorization with some formula ( that uses only rational operations and radicals) that works for any quintic.

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    $\begingroup$ 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. $\endgroup$ – Peter Jul 2 '16 at 12:35
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    $\begingroup$ In fact, a random equation of degree $5$ or higher is very likely to be non-solvable by radicals. $\endgroup$ – Peter Jul 2 '16 at 12:37
  • $\begingroup$ @Peter: I modified a bit 'my answer. It's better now? $\endgroup$ – Emilio Novati Jul 2 '16 at 13:01
  • $\begingroup$ @Peter From Wikipedia: "In principle, it could be that the equations of the fifth degree could be split in several types and, for each one of these types, there could be some algebraic solution valid within that type. Or, as Ian Stewart wrote, “for all that Abel's methods could prove, every particular quintic equation might be soluble, with a special formula for each equation.” However, this is not so, but this impossibility lies outside the scope of the Abel–Ruffini theorem and is part of the Galois theory." (my italics) Perhaps you could comment on this. $\endgroup$ – MathematicsStudent1122 Jul 2 '16 at 13:05
  • $\begingroup$ @MathematicsStudent1122 Seems , I have misinterpreted the theorem. What should I comment ? $\endgroup$ – Peter Jul 2 '16 at 13:43

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