Viète's Formulas

Theorem
Let
 * $P \left({x}\right) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$

be a polynomial of degree $n$ over a ring $R$.

Suppose $P$ can be expressed as:
 * $\displaystyle P \left({x}\right) = a_n \prod_{k \mathop = 1}^n \left({x - z_k}\right)$

where $z_1, \ldots, z_k \in R$.

That is, $z_1, \ldots, z_k$ are roots of $P$, not guaranteed to be unique in general.

Then:
 * $\displaystyle a_{n-k} = \left({-1}\right)^k a_n \sum_{1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n} z_{i_1} \dotsm z_{i_k}$

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

Listed explicitly, supposing $a_n$ is invertible:


 * $z_1 + z_2 + \cdots + z_n = -a_{n-1} / a_n$
 * $z_1 z_2 + \cdots + z_1 z_n + z_2 z_3 + \cdots + z_2 z_n + \cdots + z_{n-1} z_n = a_{n-2} / a_n$
 * $\cdots$
 * $z_1 z_2 \cdots z_n = \left({-1}\right)^n a_0 / a_n$

Proof
Note that the indexing $1 \le i_1 \le \cdots i_k \le n$ represents all possible subsets of $\left\{ {1, 2, \dotsc, n}\right \}$ of size $k$ up to order.

It follows from Product of Sums: Corollary that $P \left({x}\right)$ foils as a sum of powers $x^k$ with coefficients as sums of all products of elements of subsets of $\left\{ {z_1, \dotsc, z_n}\right\}$ of complementary size $n-k$, hence equating pairwise with the original coefficients of $P \left({x}\right)$ obtains Viète's Formulas.