Difference between revisions of "Vieta's formulas"

Revision as of 14:58, 13 July 2024

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 many math contests/tournaments.

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 $j^{\text{th}}$ elementary symmetric polynomial of the roots.

Vieta’s formulas then state that $s_1 = r_1 + r_2 + \cdots + r_n = - \frac{a_{n-1}}{a_n}$ $s_2 = r_1r_2 + r_2r_3 + \cdots + r_{n-1}r_n = \frac{a_{n-2}}{a_n}$ $\vdots$ $s_n = r_1r_2r_3 \cdots r_n = (-1)^n \frac{a_0}{a_n}.$ This can be compactly summarized as $s_j = (-1)^j \frac{a_{n-j}}{a_n}$ for some $j$ such that $1 \leq j \leq n$ .

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 Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.

When expanding the factorization of $P(x)$ , each term is generated by a series of $n$ choices of whether to include $x$ or the negative root $-r_{i}$ from every factor $(x-r_{i})$ . Consider all the expanded terms of the polynomial with degree $n-j$ ; they are formed by multiplying a choice of $j$ negative roots, making the remaining $n-j$ choices in the product $x$ , and finally multiplying by the constant $a_n$ .

Note that adding together every multiplied choice of $j$ negative roots yields $(-1)^j s_j$ . Thus, when we expand $P(x)$ , 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 $s_j = (-1)^j a_{n-j}/a_n$ , which completes the proof. $\square$

Problems

Here are some problems with solutions that utilize Vieta's quadratic formulas:

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-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 difference of the polynomial's coefficients.
+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.
-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.
+It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments.
 == 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</math>th elementary symmetric polynomial of the roots. Vietas 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_1r_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 written as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>0<j \leq n</math>
+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 these formulas by expanding this polynomial. When expanding this polynomial, every term is generated by <math>n</math> choices whether to include <math>x</math> or <math>-r_{n-j}</math> from any factor <math>(x-r_{n-j})</math>.
+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.
-Consider all the expanded terms of <math>P(x)</math> with degree <math>j</math>; they are formed by choosing <math>j</math> of the negative roots, then by making the remaining <math>n-j</math> choices <math>x</math>. Thus, every term is equal to a product of <math>j</math> of the negative roots multiplied by <math>x_{n-j}</math>. If one factors out <math>(-1^{j})x_{n-j}</math>, we are left with the <math>j</math>th elementary symmetric polynomial of the roots. Thus, when expanding <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>
+When expanding the 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 that test knowledge of Vieta's formulas.
+Here are some problems with solutions that utilize Vieta's quadratic formulas:
 === Introductory ===
@@ Line 18: / Line 23: @@
 * [[2007 AMC 12A Problems/Problem 21 | 2007 AMC 12A Problem 21]]
 * [[2010 AMC 10A Problems/Problem 21 | 2010 AMC 10A Problem 21]]
+* [[2003 AMC 10A Problems/Problem 18 | 2003 AMC 10A Problem 18]]
 === Intermediate ===
 * [[2017 AMC 12A Problems/Problem 23 | 2017 AMC 12A Problem 23]]
+* [[2003 AIME II Problems/Problem 9 | 2003 AIME II Problem 9]]
 * [[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 ==

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Difference between revisions of "Vieta's formulas"

Revision as of 14:58, 13 July 2024

Contents

Statement

Proof

Problems

Introductory

Intermediate

See also