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 difference 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 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 $j$ th elementary symmetric polynomial of the roots. Vietas 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 written as $s_j = (-1)^j \frac{a_{n-j}}{a_n}$ for some $j$ such that $0<j \leq n$

Proof

Let all terms be defined as above. By the factor theorem, $P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)$ ; we will then prove these formulas by expanding this polynomial. When expanding this polynomial, every term is generated by $n$ choices whether to include $x$ or $-r_{n-j}$ from any factor $(x-r_{n-j})$ .

Consider all the expanded terms of $P(x)$ with degree $j$ ; they are formed by choosing $j$ of the negative roots, then by making the remaining $n-j$ choices $x$ . Thus, every term is equal to a product of $j$ of the negative roots multiplied by $x_{n-j}$ . If one factors out $(-1^{j})x_{n-j}$ , we are left with the $j$ th elementary symmetric polynomial of the roots. Thus, when expanding this product, the coefficient of $x_{n-j}$ is equal to $(-1)^j a_n s_j$ . However, we defined the coefficient of $x_{n-j}$ to be $a_{n-j}$ . Thus, $(-1)^j a_n s_j = a_{n-j}$ , or $s_j = (-1)^j a_{n-j}/a_n$ , as required. $\square$