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Vieta's formulas

## Summary

In mathematics, Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots.[1] They are named after François Viète (more commonly referred to by the Latinised form of his name, "Franciscus Vieta").

## Basic formulas

Any general polynomial of degree n ${\displaystyle P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}$  (with the coefficients being real or complex numbers and an ≠ 0) has n (not necessarily distinct) complex roots r1, r2, ..., rn by the fundamental theorem of algebra. Vieta's formulas relate the polynomial coefficients to signed sums of products of the roots r1, r2, ..., rn as follows:

 ${\displaystyle {\begin{cases}r_{1}+r_{2}+\dots +r_{n-1}+r_{n}=-{\dfrac {a_{n-1}}{a_{n}}}\\[1ex](r_{1}r_{2}+r_{1}r_{3}+\cdots +r_{1}r_{n})+(r_{2}r_{3}+r_{2}r_{4}+\cdots +r_{2}r_{n})+\cdots +r_{n-1}r_{n}={\dfrac {a_{n-2}}{a_{n}}}\\[1ex]{}\quad \vdots \\[1ex]r_{1}r_{2}\cdots r_{n}=(-1)^{n}{\dfrac {a_{0}}{a_{n}}}.\end{cases}}}$ (*)

Vieta's formulas can equivalently be written as ${\displaystyle \sum _{1\leq i_{1}  for k = 1, 2, ..., n (the indices ik are sorted in increasing order to ensure each product of k roots is used exactly once).

The left-hand sides of Vieta's formulas are the elementary symmetric polynomials of the roots.

Vieta's system (*) can be solved by Newton's method through an explicit simple iterative formula, the Durand-Kerner method.

## Generalization to rings

Vieta's formulas are frequently used with polynomials with coefficients in any integral domain R. Then, the quotients ${\displaystyle a_{i}/a_{n}}$  belong to the field of fractions of R (and possibly are in R itself if ${\displaystyle a_{n}}$  happens to be invertible in R) and the roots ${\displaystyle r_{i}}$  are taken in an algebraically closed extension. Typically, R is the ring of the integers, the field of fractions is the field of the rational numbers and the algebraically closed field is the field of the complex numbers.

Vieta's formulas are then useful because they provide relations between the roots without having to compute them.

For polynomials over a commutative ring that is not an integral domain, Vieta's formulas are only valid when ${\displaystyle a_{n}}$  is not a zero-divisor and ${\displaystyle P(x)}$  factors as ${\displaystyle a_{n}(x-r_{1})(x-r_{2})\dots (x-r_{n})}$ . For example, in the ring of the integers modulo 8, the quadratic polynomial ${\displaystyle P(x)=x^{2}-1}$  has four roots: 1, 3, 5, and 7. Vieta's formulas are not true if, say, ${\displaystyle r_{1}=1}$  and ${\displaystyle r_{2}=3}$ , because ${\displaystyle P(x)\neq (x-1)(x-3)}$ . However, ${\displaystyle P(x)}$  does factor as ${\displaystyle (x-1)(x-7)}$  and also as ${\displaystyle (x-3)(x-5)}$ , and Vieta's formulas hold if we set either ${\displaystyle r_{1}=1}$  and ${\displaystyle r_{2}=7}$  or ${\displaystyle r_{1}=3}$  and ${\displaystyle r_{2}=5}$ .

## Example

Vieta's formulas applied to quadratic and cubic polynomials:

The roots ${\displaystyle r_{1},r_{2}}$  of the quadratic polynomial ${\displaystyle P(x)=ax^{2}+bx+c}$  satisfy ${\displaystyle r_{1}+r_{2}=-{\frac {b}{a}},\quad r_{1}r_{2}={\frac {c}{a}}.}$

The first of these equations can be used to find the minimum (or maximum) of P; see Quadratic equation § Vieta's formulas.

The roots ${\displaystyle r_{1},r_{2},r_{3}}$  of the cubic polynomial ${\displaystyle P(x)=ax^{3}+bx^{2}+cx+d}$  satisfy ${\displaystyle r_{1}+r_{2}+r_{3}=-{\frac {b}{a}},\quad r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}={\frac {c}{a}},\quad r_{1}r_{2}r_{3}=-{\frac {d}{a}}.}$

## Proof

### Direct proof

Vieta's formulas can be proved by expanding the equality ${\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}=a_{n}(x-r_{1})(x-r_{2})\cdots (x-r_{n})}$  (which is true since ${\displaystyle r_{1},r_{2},\dots ,r_{n}}$  are all the roots of this polynomial), multiplying the factors on the right-hand side, and identifying the coefficients of each power of ${\displaystyle x.}$

Formally, if one expands ${\displaystyle (x-r_{1})(x-r_{2})\cdots (x-r_{n}),}$  the terms are precisely ${\displaystyle (-1)^{n-k}r_{1}^{b_{1}}\cdots r_{n}^{b_{n}}x^{k},}$  where ${\displaystyle b_{i}}$  is either 0 or 1, accordingly as whether ${\displaystyle r_{i}}$  is included in the product or not, and k is the number of ${\displaystyle r_{i}}$  that are included, so the total number of factors in the product is n (counting ${\displaystyle x^{k}}$  with multiplicity k) – as there are n binary choices (include ${\displaystyle r_{i}}$  or x), there are ${\displaystyle 2^{n}}$  terms – geometrically, these can be understood as the vertices of a hypercube. Grouping these terms by degree yields the elementary symmetric polynomials in ${\displaystyle r_{i}}$  – for xk, all distinct k-fold products of ${\displaystyle r_{i}.}$

As an example, consider the quadratic ${\displaystyle f(x)=a_{2}x^{2}+a_{1}x+a_{0}=a_{2}(x-r_{1})(x-r_{2})=a_{2}(x^{2}-x(r_{1}+r_{2})+r_{1}r_{2}).}$

Comparing identical powers of ${\displaystyle x}$ , we find ${\displaystyle a_{2}=a_{2}}$ , ${\displaystyle a_{1}=-a_{2}(r_{1}+r_{2})}$  and ${\displaystyle a_{0}=a_{2}(r_{1}r_{2})}$ , with which we can for example identify ${\displaystyle r_{1}+r_{2}=-a_{1}/a_{2}}$  and ${\displaystyle r_{1}r_{2}=a_{0}/a_{2}}$ , which are Vieta's formula's for ${\displaystyle n=2}$ .

### Proof by mathematical induction

Vieta's formulas can also be proven by induction as shown below.

Inductive hypothesis:

Let ${\displaystyle {P(x)}}$  be polynomial of degree ${\displaystyle n}$ , with complex roots ${\displaystyle {r_{1}},{r_{2}},{\dots },{r_{n}}}$  and complex coefficients ${\displaystyle a_{0},a_{1},\dots ,a_{n}}$  where ${\displaystyle {a_{n}}\neq 0}$ . Then the inductive hypothesis is that${\displaystyle {P(x)}={a_{n}}{x^{n}}+{{a_{n-1}}{x^{n-1}}}+{\cdots }+{{a_{1}}{x}}+{{a}_{0}}={{a_{n}}{x^{n}}}-{a_{n}}{({r_{1}}+{r_{2}}+{\cdots }+{r_{n}}){x^{n-1}}}+{\cdots }+{{(-1)^{n}}{(a_{n})}{({r_{1}}{r_{2}}{\cdots }{r_{n}})}}}$

Base case, ${\displaystyle n=2}$  (quadratic):

Let ${\displaystyle {a_{2}},{a_{1}}}$  be coefficients of the quadratic and ${\displaystyle a_{0}}$ be the constant term. Similarly, let ${\displaystyle {r_{1}},{r_{2}}}$  be the roots of the quadratic:${\displaystyle {a_{2}x^{2}}+{a_{1}x}+a_{0}={a_{2}}{(x-r_{1})(x-r_{2})}}$ Expand the right side using distributive property:${\displaystyle {a_{2}x^{2}}+{a_{1}x}+a_{0}={a_{2}}{({x^{2}}-{r_{1}x}-{r_{2}x}+{r_{1}}{r_{2}})}}$ Collect like terms:${\displaystyle {a_{2}x^{2}}+{a_{1}x}+a_{0}={a_{2}}{({x^{2}}-{({r_{1}}+{r_{2}}){x}}+{r_{1}}{r_{2}})}}$ Apply distributive property again:${\displaystyle {a_{2}x^{2}}+{a_{1}x}+a_{0}={{a_{2}}{x^{2}}-{{a_{2}}({r_{1}}+{r_{2}}){x}}+{a_{2}}{({r_{1}}{r_{2}})}}}$ The inductive hypothesis has now been proven true for ${\displaystyle n=2}$ .

Induction step:

Assuming the inductive hypothesis holds true for all ${\displaystyle n\geqslant 2}$ , it must be true for all ${\displaystyle n+1}$ .${\displaystyle {P(x)}={a_{n+1}}{x^{n+1}}+{{a_{n}}{x^{n}}}+{\cdots }+{{a_{1}}{x}}+{{a}_{0}}}$ By the factor theorem, ${\displaystyle {(x-r_{n+1})}}$  can be factored out of ${\displaystyle P(x)}$  leaving a 0 remainder. Note that the roots of the polynomial in the square brackets are ${\displaystyle r_{1},r_{2},\cdots ,r_{n}}$ :${\displaystyle {P(x)}={(x-r_{n+1})}{[{\frac {{a_{n+1}}{x^{n+1}}+{{a_{n}}{x^{n}}}+{\cdots }+{{a_{1}}{x}}+{{a}_{0}}}{x-r_{n+1}}}]}}$ Factor out ${\displaystyle a_{n+1}}$ , the leading coefficient ${\displaystyle P(x)}$ , from the polynomial in the square brackets:${\displaystyle {P(x)}={(a_{n+{1}})}{(x-r_{n+1})}{[{\frac {{x^{n+1}}+{\frac {{a_{n}}{x^{n}}}{(a_{n+{1}})}}+{\cdots }+{{\frac {a_{1}}{(a_{n+{1}})}}{x}}+{\frac {a_{0}}{(a_{n+{1}})}}}{x-r_{n+1}}}]}}$ For simplicity sake, allow the coefficients and constant of polynomial be denoted as ${\displaystyle \zeta }$ :${\displaystyle P(x)={(a_{n+1})}{(x-r_{n+1})}{[{x^{n}}+{\zeta _{n-1}x^{n-1}}+{\cdots }+{\zeta _{0}}]}}$ Using the inductive hypothesis, the polynomial in the square brackets can be rewritten as:${\displaystyle P(x)={(a_{n+1})}{(x-r_{n+1})}{[{x^{n}}-{({r_{1}}+{r_{2}}+{\cdots }+{r_{n}}){x^{n-1}}}+{\cdots }+{{(-1)^{n}}{({r_{1}}{r_{2}}{\cdots }{r_{n}})}}]}}$ Using distributive property:${\displaystyle P(x)={(a_{n+1})}{({x}{[{x^{n}}-{({r_{1}}+{r_{2}}+{\cdots }+{r_{n}}){x^{n-1}}}+{\cdots }+{{(-1)^{n}}{({r_{1}}{r_{2}}{\cdots }{r_{n}})}}]}{-r_{n+1}}{[{x^{n}}-{({r_{1}}+{r_{2}}+{\cdots }+{r_{n}}){x^{n-1}}}+{\cdots }+{{(-1)^{n}}{({r_{1}}{r_{2}}{\cdots }{r_{n}})}}]})}}$ After expanding and collecting like terms:{\displaystyle {\begin{aligned}{P(x)}={{a_{n+1}}{x^{n+1}}}-{a_{n+1}}{({r_{1}}+{r_{2}}+{\cdots }+{r_{n}}+{r_{n+1}}){x^{n}}}+{\cdots }+{{(-1)^{n+1}}{({r_{1}}{r_{2}}{\cdots }{r_{n}}{r_{n+1}})}}\\\end{aligned}}} The inductive hypothesis holds true for ${\displaystyle n+1}$ , therefore it must be true ${\displaystyle \forall n\in \mathbb {N} }$

Conclusion:${\displaystyle {a_{n}}{x^{n}}+{{a_{n-1}}{x^{n-1}}}+{\cdots }+{{a_{1}}{x}}+{{a}_{0}}={{a_{n}}{x^{n}}}-{a_{n}}{({r_{1}}+{r_{2}}+{\cdots }+{r_{n}}){x^{n-1}}}+{\cdots }+{{(-1)^{n}}{({r_{1}}{r_{2}}{\cdots }{r_{n}})}}}$ By dividing both sides by ${\displaystyle a_{n}}$ , it proves the Vieta's formulas true.

## History

As reflected in the name, the formulas were discovered by the 16th-century French mathematician François Viète, for the case of positive roots.

In the opinion of the 18th-century British mathematician Charles Hutton, as quoted by Funkhouser,[2] the general principle (not restricted to positive real roots) was first understood by the 17th-century French mathematician Albert Girard:

...[Girard was] the first person who understood the general doctrine of the formation of the coefficients of the powers from the sum of the roots and their products. He was the first who discovered the rules for summing the powers of the roots of any equation.