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.

  • 2
    $\begingroup$ I like the rational root test. By Gauss' Lemma it finds all the integer roots as well. $\endgroup$ – Eric Towers Mar 20 '14 at 18:57
  • 2
    $\begingroup$ 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: math.stackexchange.com/questions/562694/… $\endgroup$ – Ron Gordon Mar 20 '14 at 18:59
  • $\begingroup$ 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. $\endgroup$ – TheBluegrassMathematician Mar 20 '14 at 19:05
  • 5
    $\begingroup$ @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.) $\endgroup$ – Eric Towers Mar 20 '14 at 19:12
  • 4
    $\begingroup$ @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. $\endgroup$ – Mark Bennet Mar 20 '14 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.