Difference between revisions of "Vieta's formulas"
Etmetalakret (talk | contribs) |
Etmetalakret (talk | contribs) |
||
Line 15: | Line 15: | ||
=== Introductory === | === Introductory === | ||
− | * [[2010 AMC 10A Problems/Problem 21 | | + | * [[2005 AMC 12B Problems/Problem 12 | 2005 AMC 12B Problem 12]] |
+ | * [[2007 AMC 12A Problems/Problem 21 | 2007 AMC 12A Problem 21]] | ||
+ | * [[2010 AMC 10A Problems/Problem 21 | 2010 AMC 10A Problem 21]] | ||
=== Intermediate === | === Intermediate === | ||
* [[2017 AMC 12A Problems/Problem 23 | 2017 AMC 12A Problem 23]] | * [[2017 AMC 12A Problems/Problem 23 | 2017 AMC 12A Problem 23]] | ||
+ | * [[2008 AIME II Problems/Problem 7 | 2008 AIME II Problem 7]] | ||
== See also == | == See also == |
Revision as of 17:54, 5 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 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 be any polynomial with complex coefficients with roots
, and let
be the
th elementary symmetric polynomial of the roots. Vietas formulas then state that
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 this product, the coefficient of
is equal to
. However, we defined the coefficient of
to be
. Thus,
, or
, as required.
Problems
Here are some problems that test knowledge of Vieta's formulas.