# Finding polynomial roots using symmetries

Given a quartic polynomial $$x^4+a x^3+bx^2+cx+1$$ where $a,b,$ and $c$ are in general complex coefficients, is there a way (say Vieta's formula) that let's me find all three other roots, given that I already know one of them? For example, I know that if I've given three roots $x_1, x_2, x_3$ then the fourth one is given by $x_4 = (x_1 x_2 x_3)^{-1}$. Using another example, suppose $a = c$. Then I know that if $x_0$ is a root, so is $x_0^{-1}$.

The corresponding question in the case where we dealing we a quadratic equation has a simple answer. That is, if we are given the quadratic equation $$\alpha x^2+ \beta x+\gamma$$ as well as a single root $x_0$, then we know the other root is $x_0^{-1} \frac{\gamma}{\alpha}$.

But is there such a nice formula for quartics? I feel like the answer is yes and that the answer lies somewhere in group/Galois theory (for example, in the quadratic case, the mapping $x_0 \to x^{-1}_0 \frac{\gamma}{\alpha}$ is an involution and along with the identity mapping forms the group of order two). Any help is appreciated!

if you know one root then you can set $$x^4+ax^3+bx^2+cx+1=(x-x_1)(Ax^3+Bx^2+Cx+D)$$