|
|
| (12 intermediate revisions by 10 users not shown) |
| Line 1: |
Line 1: |
| − | 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 polynomial | elementary symmetric polynomials]] of its roots can be easily expressed as a ratio between two of the polynomial's coefficients.
| + | #REDIRECT [[Vieta's Formulas]] |
| | | | |
| − | It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many mathematics contests.
| + | {{delete|no content}} |
| − | | |
| − | == Statement ==
| |
| − | Let <math>P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0</math> be any polynomial with [[Complex number | complex]] coefficients with roots <math>r_1, r_2, \ldots , r_n</math>, and let <math>s_j</math> be the <math>j^{\text{th}}</math> elementary symmetric polynomial of the roots.
| |
| − | | |
| − | Vieta’s formulas then state that <cmath>s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}</cmath> <cmath>s_2 = r_1r_2 + r_2r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}</cmath> <cmath>\vdots</cmath> <cmath>s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.</cmath> This can be compactly summarized as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>1 \leq j \leq n</math>.
| |
| − | | |
| − | == Proof ==
| |
| − | 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>. We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.
| |
| − | | |
| − | When expanding this factorization of <math>P(x)</math>, each term is generated by a series of <math>n</math> choices of whether to include <math>x</math> or the negative root <math>-r_{i}</math> from every factor <math>(x-r_{i})</math>. Consider all the expanded terms of the polynomial with degree <math>n-j</math>; they are formed by multiplying a choice of <math>j</math> negative roots, making the remaining <math>n-j</math> choices in the product <math>x</math>, and finally multiplying by the constant <math>a_n</math>.
| |
| − | | |
| − | Note that adding together every multiplied choice of <math>j</math> negative roots yields <math>(-1)^j s_j</math>. Thus, when we expand <math>P(x)</math>, the coefficient of <math>x_{n-j}</math> is equal to <math>(-1)^j a_n s_j</math>. However, we defined the coefficient of <math>x^{n-j}</math> to be <math>a_{n-j}</math>. Thus, <math>(-1)^j a_n s_j = a_{n-j}</math>, or <math>s_j = (-1)^j a_{n-j}/a_n</math>, which completes the proof. <math>\square</math>
| |
| − | | |
| − | == Problems ==
| |
| − | Here are some problems with solutions that utilize Vieta's formulas.
| |
| − | | |
| − | === Introductory ===
| |
| − | * [[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 ===
| |
| − | * [[2017 AMC 12A Problems/Problem 23 | 2017 AMC 12A Problem 23]]
| |
| − | * [[2008 AIME II Problems/Problem 7 | 2008 AIME II Problem 7]]
| |
| − | * [[2021 Fall AMC 12A Problems/Problem 23 | 2021 Fall AMC 12A Problem 23]]
| |
| − | | |
| − | == See also ==
| |
| − | * [[Polynomial]]
| |
| − | | |
| − | [[Category:Algebra]]
| |
| − | [[Category:Polynomials]]
| |
| − | [[Category:Theorems]]
| |
