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I know that through the Abel Ruffini Theorem the general solution to a polynomial of degree five or more cannot be found explicitly. But are there are any other ways to find the roots of such a polynomial besides exhaustive methods like Newton's Method? The question of finding such roots arose when trying to find the the y maximum of the unit lemniscate.

TheBluegrassMathematician
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    I like the rational root test. By Gauss' Lemma it finds all the integer roots as well. – Eric Towers Mar 20 '14 at 18:57
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    If there's a bunch of symmetry then there are some simple ways to deduce such roots; I do exactly this in evaluating the following integral: http://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1/565626#565626 – Ron Gordon Mar 20 '14 at 18:59
  • The for this specific polynomial $x^8 -2x^6 +3x^4 -2x^2 +1=0$, the test does give me rational zeroes but I don't think these are the ones I need. I'm trying to find the maximum y coordinate of the unit lemniscate so letting x=1 would not get me the correct answer because the maximum y occurs between x=0 and x=1. – TheBluegrassMathematician Mar 20 '14 at 19:05
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    @RyanMcGaha: Under the transform $x^2 \rightarrow u$ this is a fourth degree polynomial, so Abel-Ruffini is not an obstruction. (Additionally, the polynomial's roots are various sixth roots of -1.) – Eric Towers Mar 20 '14 at 19:12
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    @RyanMcGaha It would be useful if you put the polynomial in the question, because the one you have is a symmetric polynomial of degree $4$ in $y=x^2$, which can be reduced to a quadratic in $y+\frac 1y$. In short, it can be solved in a standard way by solving a chain of three quadratic equations. If you put it in the question, people will be able to show you the steps you need to take. – Mark Bennet Mar 20 '14 at 19:15
  • Oh wow. I did not see that. I knew the roots were plus or minus the sixth root of i. I used wolfram alpha but I wanted to know how to get there not just the answer and it could not provide the step by step solution. I'll try this substitution. Thank you. – TheBluegrassMathematician Mar 20 '14 at 19:15

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