Difference between revisions of "Vieta's formulas"
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| − | Let all terms be defined as above. By the [[factor theorem]], <math>P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)</math>. When we expand this polynomial, each term is generated by <math>n</math> choices of whether to include <math>x</math> or <math>-r_{n-i}</math> from any factor <math>(x-r_{n-i})</math>. We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients. | + | Let all terms be defined as above. By the [[factor theorem]], <math>P(x) = a_n (x-r_1)(x-r_2) \cdots (x-r_n)</math>. When we expand this polynomial, each term is generated by the <math>n</math> choices of whether to include <math>x</math> or <math>-r_{n-i}</math> from any factor <math>(x-r_{n-i})</math>. 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 <math>P(x)</math> with degree <math>n-j</math>; they are formed by choosing <math>j</math> of the negative roots, making the remaining <math>n-j</math> choices <math>x</math>, and finally multiplied by the constant <math>a_n</math>. We note that when we multiply <math>j</math> of the negative roots, we get <math>(-1)^j\cdot s_j</math>. | Consider all the expanded terms of <math>P(x)</math> with degree <math>n-j</math>; they are formed by choosing <math>j</math> of the negative roots, making the remaining <math>n-j</math> choices <math>x</math>, and finally multiplied by the constant <math>a_n</math>. We note that when we multiply <math>j</math> of the negative roots, we get <math>(-1)^j\cdot s_j</math>. | ||
Revision as of 19:52, 6 November 2021
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.
Vieta’s 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, 
. When we expand this polynomial, each term is generated by the 
choices of whether to include 
or 
from any factor 
. 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 
with degree 
; they are formed by choosing of the negative roots, making the remaining choices , and finally multiplied by the constant 
. We note that when we multiply of the negative roots, we get 
.
So in mathematical terms, when we expand , 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.
