# Vieta's Formulas

(Redirected from Vieta’s Formulae)

Vieta's Formulas, otherwise called Viète's Laws, are a set of equations relating the roots and the coefficients of polynomials.

## Introduction

Vieta's Formulas were discovered by the French mathematician François Viète. 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 $x^2+ax+b=0$ with solutions $p$ and $q$, then we know that we can factor it as: $x^2+ax+b=(x-p)(x-q)$

(Note that the first term is $x^2$, not $ax^2$.) Using the distributive property to expand the right side we now have $x^2+ax+b=x^2-(p+q)x+pq$

Vieta's Formulas are often used when finding the sum and products of the roots of a quadratic in the form $ax^2 + bx +c$ with roots $r_1$ and $r_2.$ They state that: $$r_1 + r_2 = -\frac{b}{a}$$ and $$r_1 \cdot r_2 = \frac{c}{a}.$$

We know that two polynomials are equal if and only if their coefficients are equal, so $x^2+ax+b=x^2-(p+q)x+pq$ means that $a=-(p+q)$ and $b=pq$. 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 $x$ term.

A similar set of relations for cubics can be found by expanding $x^3+ax^2+bx+c=(x-p)(x-q)(x-r)$.

We can state Vieta's formulas more rigorously and generally. Let $P(x)$ be a polynomial of degree $n$, so $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $x^{i}$ is ${a}_i$ and $a_n \neq 0$. As a consequence of the Fundamental Theorem of Algebra, we can also write $P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$, where ${r}_i$ are the roots of $P(x)$. We thus have that $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).$

Expanding out the right-hand side gives us $a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.$

The coefficient of $x^k$ in this expression will be the $(n-k)$-th elementary symmetric sum of the $r_i$.

We now have two different expressions for $P(x)$. These must be equal. However, the only way for two polynomials to be equal for all values of $x$ is for each of their corresponding coefficients to be equal. So, starting with the coefficient of $x^n$, we see that $a_n = a_n$ $a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$ $a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$ $\vdots$ $a_0 = (-1)^n a_n r_1r_2\cdots r_n$

More commonly, these are written with the roots on one side and the $a_i$ on the other (this can be arrived at by dividing both sides of all the equations by $a_n$).

If we denote $\sigma_k$ as the $k$-th elementary symmetric sum, then we can write those formulas more compactly as $\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $1\le k\le {n}$. Also, $-b/a = p + q, c/a = p \cdot q$.

## Proving Vieta's Formula

Basic proof: This has already been proved earlier, but I will explain it more. If we have $x^2+ax+b=(x-p)(x-q)$, the roots are $p$ and $q$. Now expanding the left side, we get: $x^2+ax+b=x^2-qx-px+pq$. Factor out an $x$ on the right hand side and we get: $x^2+ax+b=x^2-x(p+q)+pq$ Looking at the two sides, we can quickly see that the coefficient $a$ is equal to $-(p+q)$. $p+q$ is the actual sum of roots, however. Therefore, it makes sense that $p+q= \frac{-b}{a}$. The same proof can be given for $pq=\frac{c}{a}$.

Note: If you do not understand why we must divide by $a$, try rewriting the original equation as $ax^2+bx+c=(x-p)(x-q)$ SuperJJ

## General Form

For a polynomial of the form $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ with roots $r_1,r_2,r_3,...r_n$, Vieta's formulas state that: \begin{align*} s_1&= & r_1+r_2+r_3&+\cdots+r_n & &=-\frac{a_{n-1}}{a_n} \\ 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} \\ 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} \\ & & &\vdots & & \\ s_n&= & r_1r_2r_3&\cdots r_n & &=(-1)^n\frac{a_0}{a_n}.\\ \end{align*}

These formulas are widely used in competitions, and it is best to remember that when the $n$ roots are taken in groups of $k$ (i.e. $r_1+r_2+r_3...+r_n$ is taken in groups of 1 and $r_1r_2r_3...r_n$ is taken in groups of $n$), this is equivalent to $(-1)^k\frac{a_{n-k}}{a_n}$.

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