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I have the following polynomial that I would like to obtain an analytical solution to

$$ f(x)=3x(-2x^5+7x^4-7x^3+2x+1) $$

I am not certain how to approach this, and even to test if a solution exists. I have not studied Galois theory although I see that this is relevant. It is of order 6 in $x$, but can it be reduced to a quintic if I extract $x$ in this way?

AngusTheMan
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  • Have you tried the rational root theorem? If you can't find any rational roots then I think it will not be easy. –  Aug 24 '20 at 09:32
  • https://www.wolframalpha.com/input/?i=roots+of+3x%28-2x%5E5%2B7x%5E4-7x%5E3%2B2x%2B1%29

    Given that wolfram alpha only shows approximate solutions, I wouldn't have high hopes about finding analytical solutions...

     – 5xum Aug 24 '20 at 09:35
  • Thank you for the references, is there a way I can show there is no solution for this polynomial in that case? – AngusTheMan Aug 24 '20 at 09:36
  • [Apart $x=0$]. Two complex conjugate roots + one real root; for the latter, using the inverse symbolic calculator http://wayback.cecm.sfu.ca/cgi-bin/isc/lookup?number=+1.7157858546196886328083665733767&lookup_type=simple, no match found. – Jean Marie Aug 24 '20 at 09:47

1 Answers1

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pari/gp calculation

? polgalois(-2*x^5 + 7*x^4 - 7*x^3 + 2*x + 1)
%2 = [120, -1, 1, "S5"]

shows that this polynomial has $S_5$ Galois group. Galois theory says that a polynomial with that Galois group cannot be solved using radicals.

  • I'm curious there is a way to determine if a quintic polynomial is actually solvable or not, in a pragmatic way, rather than the abstract Galois theory. If we have a quintic with some rational roots, will the polynomial has a group isomorphic to a proper subgroup of $S_5$? – kelvin hong 方 Aug 24 '20 at 09:48
  • @rain1 out of interest, does this result from Galois theory imply that a general formula (for all solutions at once) does not exist, i.e. analogous to how the quadratic formula states all (two) solutions at once? Or can specific solutions still be written in terms of roots? – pshmath0 Aug 24 '20 at 09:48
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    @kelvinhong方, The possible Galois groups for an irreducible quintic are listed in the section here "On 5 points" https://people.maths.bris.ac.uk/~matyd/GroupNames/T15.html Both S_5 and A_5 Galois groups mean a quintic is not solvable. –  Aug 24 '20 at 09:57
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    @Pixel, yeah, it's impossible to write even one of them in terms of radicals. –  Aug 24 '20 at 09:59
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    @rain1 wow I was never sure about that, so thanks for clarifying ! Then the solutions must be transcendental? – pshmath0 Aug 24 '20 at 13:10
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    No @pixel, the solutions are still algebraic numbers. But they cannot be rendered in terms of the operations we normally call "algebraic". Certain combinations of transcendental functions also give algebraic-number results and that is what happens with roots of non-raducal soluble quintics. – Oscar Lanzi Aug 24 '20 at 13:18
  • @OscarLanzi ah yes of course they are algebraic by definition since they solve the given polynomial (silly me). So it must be possible to find some (possibly radical) expression $r(t)$ where $t$ is transcendental such that $f(r(t))=0$ ? – pshmath0 Aug 24 '20 at 14:54
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    See this question for one approach based on elliptic functions. The elliptic functions are inherently transcendental and non-elementary, it is that specific combination that becomes algebraic. – Oscar Lanzi Aug 24 '20 at 15:47
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    @OscarLanzi this is awesome, thank you. – pshmath0 Aug 25 '20 at 07:34