Vieta's formulas

In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients.

It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in all mathematics contests.

Statement

Let $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ be any polynomial with complex coefficients with roots $r_1, r_2, \ldots , r_n$ , and let $s_j$ be the $j^{\text{th}}$ elementary symmetric polynomial of the roots.

Vieta’s formulas then state that $s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}$ $s_2 = r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}$ $\vdots$ $s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.$ This can be compactly summarized as $s_j = (-1)^j \frac{a_{n-j}}{a_n}$ for some $j$ such that $1 \leq j \leq n$ .

Proof

Let all terms be defined as above. By the factor theorem, $P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)$ . When we expand this polynomial, each term is generated by the $n$ choices of whether to include $x$ or $-r_{n-i}$ from any factor $(x-r_{n-i})$ . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.

Consider all the expanded terms of $P(x)$ with degree $n-j$ ; they are formed by choosing $j$ of the negative roots, making the remaining $n-j$ choices $x$ , and finally multiplied by the constant $a_n$ . We note that when we multiply $j$ of the negative roots, we get $(-1)^j\cdot s_j$ .