Viète's Formulas

Formulas
For a polynomial $$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$ with roots $$z_1, z_2, \dots, z_n$$ (not necessarily distinct),


 * $$ \sum_{1 \le i_1 < \cdots < i_k \le n} z_{i_1} \cdots z_{i_k} = (-1)^k \frac{a_{n-k}}{a_n} $$

for $$k = 1, 2, \dots, n$$. Listed explicitly,


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

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

Setting $$P(x) = a_n (x - z_1) \cdots (x- z_n)$$, which is permitted by the fundamental theorem of algebra, it follows from a corollary to the Products of Sums formula that $$P(x)$$ foils as a sum of powers $$x^k$$ with coefficients as sums of all products of elements of subsets of $$\{ z_1, \dots, z_n \}$$ of complementary size $$n-k$$, hence equating pairwise with the original coefficients of $$P(x)$$ (divided through by $$a_n$$) obtains Viète's Formulas.