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So according to Abel-Ruffini Theorem, it states that there is no algebraic solution, in the form of radicals, to general polynomials of degree $5$ or higher.

But I'm wondering if there is a way to decide whether a polynomial, such as $$x^5+14x^4+12x^3+9x+2=0$$ has roots that can be expressed in radicals or not just by having a glance at the polynomial.

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    $\begingroup$ You could compute its Galois Group to check if you can solve it in radicals. $\endgroup$ – Edward Evans Oct 6 '16 at 23:08
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    $\begingroup$ Try Wikipedia. Of course that requires a lot more than a glance. A CAS would be handy. $\endgroup$ – Robert Israel Oct 6 '16 at 23:11
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    $\begingroup$ Modulo $3$ it has a factor of degree $3$, and modulo $7$ it is irreducible, so its Galois group contains $A_5$, hence it is not solvable. That's a bit more than a glance though, and requires some knowledge of group theory. $\endgroup$ – Servaes Oct 6 '16 at 23:12
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    $\begingroup$ +1 for the good question. Unfortunately, as commenters tell you, there is no way to answer it given your level of mathematical knowledge. $\endgroup$ – Ethan Bolker Oct 6 '16 at 23:23
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    $\begingroup$ @Simple Art Yes. But none of the roots work. The problem with the rational roots theorem is if the root is... well, irrational. Like $\sqrt2$. So just the roots from the rational root theorem isn't going to decide whether the polynomial can be solved. $\endgroup$ – Frank Oct 7 '16 at 2:50
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As the others have commented, to know when a quintic (or higher) is solvable in radicals requires Galois theory. However, there is a rather simple aspect when it is not solvable that is easily understood and can be used as a litmus test.

Theorem: An irreducible equation of prime degree $p>2$ that is solvable in radicals has either $1$ or $p$ real roots.

(Irreducible, simply put, means it has no rational roots.) By sheer coincidence, the irreducible quintic you chose has $3$ real roots so, by looking at its graph, you can indeed tell at a glance that this is not solvable in radicals. Going higher, if an irreducible septic has $3$ or $5$ real roots, then you automatically know it is not solvable. And so on.

P.S. And before you ask, it does not work the other direction: if it has $1$ or $p$ real roots, it does not imply it is solvable in radicals. It is a necessary but not sufficient condition.

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  • $\begingroup$ I have studied Galois theory, but I have never come across that theorem! How is it proved? $\endgroup$ – Unit Nov 3 '16 at 15:20
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    $\begingroup$ @Unit: I believe Dummit mentions it briefly in his paper on quintics, but I know it was established before him. How to prove it, I leave to those more capable than I. :) $\endgroup$ – Tito Piezas III Nov 3 '16 at 15:24

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