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 $x^4-10x^2+1$ ?
 A: 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.
A: To use the resolvent cubic method, first construct the resolvent cubic itself.  Using your formula for the problem at hand you get
$z^3+20z^2+96z=0$
Next, and this should be one for any quartic equation with integers, check for a rational root. Whenever the rational root exisis, we can factor the resolvent cubic without going to those messy combinations of square and cube roots (or worse!) in the general cubic formula.  That's when you get relatively "simple" quartic roots.
We can do this with the Rational Root Theorem in the general case, but here, when you have only the even degree terms, there will alwaus be the rational root $z=0$.  Thus
$z(z^2+20z+96)=0$
and upon solving the remaining quadratic factor for its zeroes we ultimately get $z\in\{-12,-8,0\}$.
The next step is to render three numbers $m,n,p$ which are the square roots of $-z/4$ values obtained from the cubic equation.  For the example here we get $\sqrt{3},\sqrt{2},0$.  With these square roots in hand the roots of the quartic equation are either set A or set B below.
Set A:  $\{m+n+p, -m-n+p, -m+n-p, m-n-p\}$
Set B:  $\{-m-n-p, m+n-p, m-n+p, -m+n+p\}$
You need to check which set applies, but for the special case where $z=0$ is a root above (because $r=0$) you can drop this step as both sets then become the same.  In your example the roots from either set are $\pm\sqrt{3}\pm\sqrt{2}$ for all four sign combinations.
When there is no zero root for $z$, meaning $r$ is nonzero, one way to do the check is to add up the reciprocals of the roots from either set and see if this sum matches $-s/r$ including the sign.  If one set fails to match signs the other must work.
