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 
be any polynomial with complex coefficients with roots 
, and let 
be the 
th elementary symmetric polynomial of the roots. Vietas formulas then state that ![\[s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}\]](http://latex.artofproblemsolving.com/4/8/f/48f4bdf0997f87131541283a8db9b4b3f9b9be09.png)
![\[s_2 = r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}\]](http://latex.artofproblemsolving.com/b/4/6/b463058fe37152140aeb6a5f5b291b04020aec11.png)
![\[\vdots\]](http://latex.artofproblemsolving.com/7/6/4/76403adfdd35fb846675aaeaac92daec680a4fba.png)
![\[s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.\]](http://latex.artofproblemsolving.com/8/c/3/8c309b1382001729c054020a9b65792d6cc5800c.png)
This can be compactly written as 
for some such that 
Proof
Let all terms be defined as above. By the factor theorem, 
; we will then prove these formulas by expanding this polynomial. When expanding this polynomial, every term is generated by 
choices whether to include 
or 
from any factor 
.
Consider all the expanded terms of 
with degree ; they are formed by choosing of the negative roots, then by making the remaining 
choices . Thus, every term is equal to a product of of the negative roots multiplied by 
. If one factors out 
, we are left with the th elementary symmetric polynomial of the roots. Thus, when expanding , the coefficient of is equal to 
. However, we defined the coefficient of to be 
. Thus, 
, or 
, which completes the proof. 
Problems
Here are some problems that test knowledge of Vieta's formulas.
