I notice that any polynomial with only even exponents is always symmetric. So I believe that for any such polynomial there is an expression of half the degree (all exponents divided by 2) that gives the same positive roots, from which the negative roots can also be derived by switching the sign.

I found that for the simplest example: $$ 0 = ax^2+b $$ The reduced expression is: $$ 0=\operatorname{sign}(a)\operatorname{abs}(a)^{(1/2)}x+\operatorname{sign}(b)\operatorname{abs}(b)^{(1/2)} $$ Where if $x>=0$ the roots of the former expression are $\pm x$ and if $x<0$ there are no real roots.

This can be seen on a graph here: https://www.desmos.com/calculator/ngdgxc6jvr

But I came to this by trial and error just playing around on Desmos, and I don't really know what I'm doing. I tried the same thing with the expression: $$ 0=ax^4+bx^2+c $$ and found that it is not so simple...

Also I am here because I have a function I have to solve in the form: $$ 0=ax^6+bx^4+cx^2+d $$

So my question is, is there an established method or pattern for reducing symmetric polynomials in this way, or does anyone care to figure it out? Does anyone have such a solution for the sextic equation mentioned above?

ps: ELI5 please, I'm a code guy not a math guy.

  • 1
    $\begingroup$ If you have to solve $ax^6+bx^4+cx^2+d=0$, you can begin by using $y=x^2$ and then you can solve $ay^3+by^2+cy+d=0$, which is the well-known cubic equation. This approach should also give you some ideas for your first question as well. $\endgroup$ – abiessu Jul 14 '18 at 4:04
  • $\begingroup$ @abiessu thanks, I'll mull that over... $\endgroup$ – Byron Jul 14 '18 at 4:11
  • $\begingroup$ @abiessu Yea that solves my problem. desmos.com/calculator/izt52qokuh $\endgroup$ – Byron Jul 14 '18 at 4:27

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