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Difference between revisions of "Vieta's Formulas"

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'''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.
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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 ratio between two of the polynomial's coefficients.
  
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== Introduction ==
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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.
  
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Vieta's Formulas were discovered by the French mathematician [[François Viète]].
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== Statement ==
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Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as:
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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.
  
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<center><math>x^2+ax+b=(x-p)(x-q)</math></center>
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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>.
  
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(Note that the first term is <math>x^2</math>, not <math>ax^2</math>.) Using the distributive property to expand the right side we now have
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== Proof ==
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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>. We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.
  
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<center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center>  
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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>.
  
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Vieta's Formulas are often used when finding the sum and products of the roots of a quadratic in the form <math>ax^2 + bx +c</math> with roots <math>r_1</math> and <math>r_2.</math> They state that:
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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>
  
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<cmath>r_1 + r_2 = -\frac{b}{a}</cmath> and <cmath>r_1 \cdot r_2 = \frac{c}{a}.</cmath>
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== Problems ==
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Here are some problems with solutions that utilize Vieta's quadratic formulas:
  
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We know that two polynomials are equal if and only if their coefficients are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.
 
  
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A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.
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=== Introductory ===
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* [[2005 AMC 12B Problems/Problem 12 | 2005 AMC 12B Problem 12]]
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* [[2007 AMC 12A Problems/Problem 21 | 2007 AMC 12A Problem 21]]
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* [[2010 AMC 10A Problems/Problem 21 | 2010 AMC 10A Problem 21]]
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* [[2003 AMC 10A Problems/Problem 18 | 2003 AMC 10A Problem 18]]
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* [[2021 AMC 12A Problems/Problem 12 | 2021 AMC 12A Problem 12]]
  
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We can state Vieta's formulas more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,
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=== Intermediate ===
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where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>.  We thus have that
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* [[2017 AMC 12A Problems/Problem 23 | 2017 AMC 12A Problem 23]]
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* [[2003 AIME II Problems/Problem 9 | 2003 AIME II Problem 9]]
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* [[2008 AIME II Problems/Problem 7 | 2008 AIME II Problem 7]]
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* [[2021 Fall AMC 12A Problems/Problem 23 | 2021 Fall AMC 12A Problem 23]]
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* [[2019 AIME I Problems/Problem 10 | 2019 AIME I Problem 10]]
  
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<center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center>
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==Advanced==
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* [[2020 AIME I Problems/Problem 14 | 2020 AIME I Problem 14]]
  
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Expanding out the right-hand side gives us
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== See also ==
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* [[Polynomial]]
  
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<center><math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math></center>
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[[Category:Algebra]]
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[[Category:Polynomials]]
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The coefficient of <math> x^k </math> in this expression will be the <math>(n-k)</math>-th [[elementary symmetric sum]] of the <math>r_i</math>. 
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[[Category:Theorems]]
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We now have two different expressions for <math>P(x)</math>.  These must be equal.  However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal.  So, starting with the coefficient of <math> x^n </math>, we see that
 
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<center><math>a_n = a_n</math></center>
 
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<center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center>
 
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<center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center>
 
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<center><math>\vdots</math></center>
 
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<center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center>
 
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More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).
 
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If we denote <math>\sigma_k</math> as the <math>k</math>-th elementary symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</math>.
 
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Also, <math>-b/a = p + q, c/a = p \cdot q</math>.
 
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==Proving Vieta's Formula==
 
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Basic proof:
 
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This has already been proved earlier, but I will explain it more.
 
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If we have
 
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<math>x^2+ax+b=(x-p)(x-q)</math>, the roots are <math>p</math> and <math>q</math>.
 
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Now expanding the left side, we get: <math>x^2+ax+b=x^2-qx-px+pq</math>.
 
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Factor out an <math>x</math> on the right hand side and we get: <math>x^2+ax+b=x^2-x(p+q)+pq</math>
 
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Looking at the two sides, we can quickly see that the coefficient <math>a</math> is equal to <math>-(p+q)</math>. <math>p+q</math> is the actual sum of roots, however. Therefore, it makes sense that <math>p+q= \frac{-b}{a}</math>. The same proof can be given for <math>pq=\frac{c}{a}</math>.
 
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Note: If you do not understand why we must divide by <math>a</math>, try rewriting the original equation as <math>ax^2+bx+c=(x-p)(x-q)</math>
 
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SuperJJ
 
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==General Form==
 
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For a polynomial of the form <math>f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0</math> with roots <math>r_1,r_2,r_3,...r_n</math>, Vieta's formulas state that:
 
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<cmath>
 
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\begin{align*}
 
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s_1&= & r_1+r_2+r_3&+\cdots+r_n & &=-\frac{a_{n-1}}{a_n} \
 
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s_2&= & r_1r_2+r_1r_3+r_1r_4&+\cdots+r_{n-2}r_{n-1} & &=\phantom{-}\frac{a_{n-2}}{a_n} \
 
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s_3&= & r_1r_2r_3+r_1r_2r_4&+\cdots+r_{n-2}r_{n-1}r_n & &=-\frac{a_{n-3}}{a_n} \
 
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& & &\vdots & & \
 
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s_n&= & r_1r_2r_3&\cdots r_n & &=(-1)^n\frac{a_0}{a_n}.\
 
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\end{align*}
 
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</cmath>
 
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These formulas are widely used in competitions, and it is best to remember that when the <math>n</math> roots are taken in groups of <math>k</math> (i.e. <math>r_1+r_2+r_3...+r_n</math> is taken in groups of 1 and <math>r_1r_2r_3...r_n</math> is taken in groups of <math>n</math>), this is equivalent to <math>(-1)^k\frac{a_{n-k}}{a_n}</math>.
 

Latest revision as of 14:21, 24 June 2025

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:


Introductory

Intermediate

Advanced

See also

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