Resolvent Cubic of Quartic Polynomial I know the resolvent cubic $(z-a)(z-b)(z-c)$ of $x^4+qx^2+rx+s$ is $z^3-2qz^2+(q^2-4s)z+r^2$. I am wondering how to use this result  (by first solving the resolvent cubic?) to find the solutions to $x^4-10x^2+1$ ?

When solving an equation of the form $$ax^4+bx^2+c=0$$note it is a hidden quadratic. 
Notice it can also be written as $a(x^2)^2+b(x^2)^1+c(x^2)^0$ and so by the quadratic formula we have:
$$x^2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ and thus
$$x=\pm\sqrt{\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$$
I don't know a similar shortcut for if your equation has an $x$ term, though, but this method is far easier if it does not.
