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I'm trying to solve a problem, but it involves finding the exact roots of the octic polynomial

$$x^8+4x^7-10x^6-54x^5+9x^4+226x^3+125x^2-301x-269$$

How can I find the roots of an octic? Wolfram Alpha just gives me the rounded values. Not the exact ones.

Rodrigo de Azevedo
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Frank
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    In general, there is no closed form for the roots of a generic polynomial of degree greater than or equal to five. You might get lucky and it will factor simply, but such occurrences are rare. See the Abel-Ruffini theorem. – JMoravitz Jul 25 '16 at 19:29
  • So there is quite a large possibility that there is no way to write the roots of this polynomial in terms of radicals? – Frank Jul 25 '16 at 19:31
  • Correct. Most equations of degree five or higher do NOT have roots that can be expressed by radicals and rational numbers. –  Jul 25 '16 at 19:36
  • Rats... =( This isn't a homework assignment by the way – Frank Jul 25 '16 at 19:37
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    It is irreducible over the rationals. Not much to be done; not just a lack of rational roots, if there were anything nice about the roots, it would be possible to factor this. It does factor over the real numbers as a product of quadratics and linears, but those coefficients cannot be found either. – Will Jagy Jul 25 '16 at 19:40
  • Okay, thanks... – Frank Jul 25 '16 at 19:41
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    A bit of extra-context wouldn't hurt anyone. By chance, is that polynomial involved in some integral? – Jack D'Aurizio Jul 25 '16 at 19:44
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    Hey Jack; No... I need to find the roots of it to find $x$ in $x=\sqrt{5-\sqrt{5-\sqrt{5+\sqrt{5+x}}}}$ – Frank Jul 25 '16 at 20:01
  • Maybe it helps to rewrite it as $$x (x-1) \left( x \left( x (x+1) \left( x (x^2+4 x-9)-50 \right)+176 \right) + 301 \right) - 269 = 0 $$ – John Alexiou Jul 25 '16 at 20:25
  • @mathguy:This particular octic is solvable in radicals. – Tito Piezas III Jul 25 '16 at 20:55

1 Answers1

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The root $x=1.87178\dots$ is a root of the quadratic,

$$4x^2+2\Big(2-\sqrt{5}+\sqrt{13-4\sqrt{5}}\Big)x-\Big(7+5\sqrt{5}+\sqrt{118-50\sqrt{5}}\Big)=0$$

which can then be solved using the well-known quadratic formula.

You can factor your octic into two quartics over $\sqrt{5}$, then those quartics into four quadratics over $\sqrt{13\pm4\sqrt{5}}\,$ (like the above), but you'll need Mathematica or Maple for a quick result.

Tito Piezas III
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  • The general case $x = \sqrt{a\pm \sqrt{a\pm \sqrt{a\pm \sqrt{a\pm\dots}}}}$ with period $4$ is discussed in this post. Yours is the case $a=5$. – Tito Piezas III Jul 25 '16 at 20:59
  • Wait, but according to the post, how will you know which root to take as the answer to the infinitely nested radical? There are $16$ possible roots, and only $1$ of them are correct... – Frank Jul 26 '16 at 01:39
  • @Frank: All of them can be correct, but you have to match it properly to the sign changes you choose. For example, the exact solutions to, $$x_1 = \sqrt{5- \sqrt{5- \sqrt{5+ \sqrt{5+x_1}}}}$$ $$x_2 = -\sqrt{5+ \sqrt{5+ \sqrt{5- \sqrt{5+x_2}}}}$$ are the two roots of the quadratic given in the answer, or $x_1\approx 1.8717$, and $x_2\approx -2.7606$. Which goes with what is primarily by inspection. – Tito Piezas III Jul 26 '16 at 03:16
  • :O Okay, that helps a lot! – Frank Jul 26 '16 at 03:17
  • Strictly speaking, the quartics are not over $\mathbb Z[\sqrt5]$. As often happens when the radicand is one greater than a multiple of 4, you get algebraic integers with half-integer components, thus $\mathbb Z[\tau],\tau=(1+\sqrt5)/2$. In terms of this we identify a factor $x^4-(2\tau-3)x^3-(5\tau+2)x^2+(2\tau-1)x+(15\tau-4)$ containing the root. – Oscar Lanzi Feb 07 '22 at 16:33