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This quintic has 5 real roots, how do we find out if it is solvable and ,in that case, how to solve it? Is there a generally valid numeric approach?

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    "a generally valid numeric approach" Newton's method works for most polynomials. I'm sure there are better tools more specialised for polynomials. As for solvability, is $8.58368$ exact, or rounded from some square root or something? It seems a little out of place here. – Arthur Apr 12 '18 at 06:15
  • Define solvable. Do you mean closed form roots (no), number of real roots (yes), range of RHS such that it has $5$ real roots (yes), or ...? – dxiv Apr 12 '18 at 06:23
  • If 2.8 is an exact root, then you can divide the original equation $x^{5}-10x^{3}+20x-8.58368$ by $x-2.8$ and solve the quartic equation. – Seewoo Lee Apr 12 '18 at 07:32

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In general, a quintic cannot be solved exatly by radicals. But substiture $x\leftarrow y/10$ to arrive at $$y^5-1000 y^3+200000y = 858368$$ First, we look for rational solutions, which must be among the divisors of $858368$. After finitely many trials, we find $y=28$ ($x=2.8$). The remaining solutions hide in $$ y^4 + 28y^3 - 216y^2 - 6048y + 3065600=0,$$ which unfortunately does not factor so easily. While quartics are solvable by radicals, one would usually still prefer to resort to numerical methods to tackle them (as in, usually a quite exact numerical value helps you more than an extremely convoluted exact expression).

Hagen von Eitzen
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Hint:

Let $x=a\sin t$

$$\dfrac{a^5}{16}=\dfrac{10a^3}{20}=\dfrac{20a}5$$

$\implies a^2=8,$

$$\implies x^5-10x^3+20x=\pm?\sin5t\le?$$ for real $x$

lab bhattacharjee
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