Sum of Roots of Polynomial

Theorem
Let $P$ be the polynomial equation:
 * $a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 = 0$

such that $a_n \ne 0$.

The sum of the roots of $P$ is $- \dfrac {a_{n-1}} {a_n}$.

Proof
Let the roots of $P$ be $z_1, z_2, \ldots, z_n$.

Then $P$ can be written in factored form as:
 * $\displaystyle a_n \prod_{k \mathop = 1}^n \left({z - z_k}\right) = a_0 \left({z - z_1}\right) \left({z - z_2}\right) \cdots \left({z - z_n}\right)$

Multiplying this out, $P$ can be expressed as:
 * $a_n \left({z^n - \left({z_1 + z_2 + \cdots + z_n}\right) z^{n-1} + \cdots + \left({-1}\right)^n z_1 z_2 \cdots z_n}\right) = 0$

where the coefficients of $z^{n-2}, z^{n-3}, \ldots$ are more complicated and irrelevant.

Equating powers of $z$, it follows that:
 * $-a_n \left({z_1 + z_2 + \cdots + z_n}\right) = a_{n-1}$

from which:
 * $z_1 + z_2 + \cdots + z_n = - \dfrac {a_{n-1}} {a_n}$