Vieta's formulas

Statement

Let $R$ be a commutative ring and let $f = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 \in R[x]$. Suppose that $f = a_n (x - r_1)(x - r_2) \cdots (x - r_n)$ with $r_1, \dots, r_n \in R$. Then for $1 \le k \le n$, $$a_n \, e_k(r_1, \dots, r_n) = (-1)^k a_{n-k},$$ where $e_k$ is the $k$-th elementary symmetric polynomial.

Remarks

If $a_n$ is a unit, this says $e_k(r_1, \dots, r_n) = (-1)^k a_{n-k}/a_n$. For $x^2 + b x + c = (x - r_1)(x - r_2)$, it gives $r_1 + r_2 = -b$ and $r_1 r_2 = c$. The proof expands the product and compares coefficients. Over a field, the hypothesis says that $f$ splits with roots $r_1, \dots, r_n$, counted with multiplicity.

History

The formulas are named after the French mathematician François Viète, who derived them for positive roots.

Explanations

L1 human

Vieta's formulas say that the sum of the roots, the sum of their products in pairs, and so on, are given by the coefficients. You can know these sums without finding the roots.