# How would one go about solving a complicated polynomials? (Not necessarily complex one's, though)

Say I have an equation such as: $$x^2 + 3x^4y^2 + 3y^4x^2 + y^6 = 5x^2y^2$$ How would one go about solving such an equation? Where $y$ could be isolated in terms of $x$. I.e, $y =$ formula involving $x$.

If you want find the solutions for $y$ you have to ordinate the polynomial with the $y$ powers and in your case we have: $$y^6+3x^2y^4+y^2(3x^4-5x^2)+x^2=0$$ In general this is a polynomial of degree $n$ (in you case $n=6$) and, for $n>4$ we have not general formulas that gives a solution by means of radicals ( Abel-Ruffini theorem). In this case we can put $y^2=t$ and the equation becomes $$t^3+3x^2t^2+(3x^4-5x^2)t+x^2=0$$ That is a third degree equation with a ''parameter'' $x$. So, if we can solve such equation we can find a formula that gives $y^2=F(x)$.

Depending if you want for $y$ real or complex numbers you have to discuss such solution to find the good values of $x$ .

I don't see a simple way to solve it without using the Cardano formula.(and the discussion of solutions is not easy at all).

• Ahhh, yes, I have worked with 3rd degrees, and 4th degrees by taking the square root of the 4th degrees and what not, but it's interesting in that established methods seem to break down beyond the 4th degree, as this is exactly what seems to be happening. Thank you very much for the links to the Abel-Ruffini Theorem, and cardano formula. I'll also be sure to work through the methodology you presented here. – Jim Jam Oct 4 '15 at 13:36
• You are welcome!.... and not forget to accept if you like :) – Emilio Novati Oct 4 '15 at 13:48

Maybe a slightly easier approach is to complete the cube first as $$x^2+(x^2+y^2)^3-x^6=5x^2y^2$$ and introduce the variable $z=x^2+y^2$ to get $$0=x^2+z^3-x^6-5x^2(z-x^2)=z^3-5x^2z-x^6+5x^4+x^2$$ which is a depressed cubic (without a quadratic term) in $z$.

One can further divide the depressed cubic by $x^3$ (assuming $x\ne 0$) and make the change $t=\frac{z}{x}$ to get $$f(x,t)=t^3-5t-x^3+5x+\frac{1}{x}=0$$

P.S. It is easy to show that the only real solution to the original equation is $x=y=0$. For example, one can minimize the function $f(x,t)$ for $x>0$, $t\ge x$ and see that the minimum is positive.