Sum of Roots of Polynomial

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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 1

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

Then $P$ can be written in factored form as:

$\displaystyle a_n \prod_{k \mathop = 1}^n \paren {z - z_k} = a_n \paren {z - z_1} \paren {z - z_2} \cdots \paren {z - z_n}$

Multiplying this out, $P$ can be expressed as:

$a_n \paren {z^n - \paren {z_1 + z_2 + \cdots + z_n} z^{n - 1} + \cdots + \paren {-1}^n z_1 z_2 \cdots z_n} = 0$

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

Equating powers of $z$, it follows that:

$-a_n \paren {z_1 + z_2 + \cdots + z_n} = a_{n - 1}$

from which:

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

$\blacksquare$


Proof 2

From Viète's Formulas:

$\displaystyle a_{n - k} = \paren {-1}^k a_n \sum_{1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n} z_{i_1} \dotsm z_{i_k}$

for $k = 1, 2, \ldots, n$.

The result follows for $k = 1$.

$\blacksquare$