Polynomials: Vieta's Formulas - Generalized

We move to cubic and quartic equations in our exploration of the coefficients and roots of the polynomial, and discover that we can derive Vieta's formulas for a polynomial of any degree using an interesting combinatorial perspective.
Exercise: Try to find the coefficient of the x^2 term in a polynomial of degree 6 in terms of the roots (r1, r2, r3, r4, r5, r6).

[thomasolson]

Spoiler alert.
r[1]=e^(i*θ[1])
r[2]=e^(i*θ[2])
r[3]=e^(i*θ[3])
​Me thinks that's algebra in a nutshell. The theta's probably have some kind of relationship. I think they all add up to pi. And you can probably cancel the 'i' and go hyperbolic (arctan(i*3/5)). I'm just a moron though. I only completed high school. I might even be autistic. I'm definitely weird enough.