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Difference between revisions of "Vieta's formulas"

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== Problems ==
 
== Problems ==
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Here are some problems with solutions that utilize Vieta's formulas:)
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Here are some problems with solutions that utilize Vieta's formulas :)
  
  

Revision as of 21:12, 26 January 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 mathematics contests.

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 this 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 formulas :)


Introductory

Intermediate

See also

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