Vieta's formulas

In mathematics, Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. 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 $$P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$$ (with the coefficients being real or complex numbers and $a_{n} ≠ 0$) has $n$ (not necessarily distinct) complex roots $r_{1}, r_{2}, ..., r_{n}$ by the fundamental theorem of algebra. Vieta's formulas relate the polynomial coefficients to signed sums of products of the roots $r_{1}, r_{2}, ..., r_{n}$ as follows:

Vieta's formulas can equivalently be written as $$\sum_{1\le i_1 < i_2 < \cdots < i_k\le n} \left(\prod_{j = 1}^k r_{i_j}\right)=(-1)^k\frac{a_{n-k}}{a_n}$$ for $k = 1, 2, ..., n$ (the indices $i_{k}$ 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 $$. Then, the quotients $$a_i/a_n$$ belong to the field of fractions of $R$ (and possibly are in $R$ itself if $$a_n$$ happens to be invertible in $R$) and the roots $$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 $$a_n$$ is not a zero-divisor and $$P(x)$$ factors as $$a_n(x-r_1)(x-r_2)\dots(x-r_n)$$. For example, in the ring of the integers modulo 8, the quadratic polynomial $$P(x) = x^2-1$$ has four roots: 1, 3, 5, and 7. Vieta's formulas are not true if, say, $$r_1=1$$ and $$r_2=3$$, because $$P(x)\neq (x-1)(x-3)$$. However, $$P(x)$$ does factor as $$(x-1)(x-7)$$ and also as $$(x-3)(x-5)$$, and Vieta's formulas hold if we set either $$r_1=1$$ and $$r_2=7$$ or $$r_1=3$$ and $$r_2=5$$.

Example
Vieta's formulas applied to quadratic and cubic polynomials:

The roots $$r_1, r_2$$ of the quadratic polynomial $$P(x) = ax^2 + bx + c$$ satisfy $$ r_1 + r_2 = -\frac{b}{a}, \quad r_1 r_2 = \frac{c}{a}.$$

The first of these equations can be used to find the minimum (or maximum) of $P$; see.

The roots $$r_1, r_2, r_3$$ of the cubic polynomial $$P(x) = ax^3 + bx^2 + cx + d$$ satisfy $$ r_1 + r_2 + r_3 = -\frac{b}{a}, \quad r_1 r_2 + r_1 r_3 + r_2 r_3 = \frac{c}{a}, \quad r_1 r_2 r_3 = -\frac{d}{a}.$$

Proof
Vieta's formulas can be proved by expanding the equality $$a_n x^n + a_{n-1}x^{n-1} +\cdots + a_1 x+ a_0 = a_n (x-r_1) (x-r_2) \cdots (x-r_n)$$ (which is true since $$r_1, r_2, \dots, r_n$$ are all the roots of this polynomial), multiplying the factors on the right-hand side, and identifying the coefficients of each power of $$x.$$

Formally, if one expands $$(x-r_1) (x-r_2) \cdots (x-r_n),$$ the terms are precisely $$(-1)^{n-k}r_1^{b_1}\cdots r_n^{b_n} x^k,$$ where $$b_i$$ is either 0 or 1, accordingly as whether $$r_i$$ is included in the product or not, and k is the number of $$r_i$$ that are included, so the total number of factors in the product is n (counting $$x^k$$ with multiplicity k) – as there are n binary choices (include $$r_i$$ or x), there are $$2^n$$ terms – geometrically, these can be understood as the vertices of a hypercube. Grouping these terms by degree yields the elementary symmetric polynomials in $$r_i$$ – for xk, all distinct k-fold products of $$r_i.$$

As an example, consider the quadratic $$f(x) = a_2x^2 + a_1x + a_0 = a_2(x - r_1)(x - r_2) = a_2(x^2 - x(r_1 + r_2) + r_1 r_2).$$

Comparing identical powers of $$x$$, we find $$a_2=a_2$$, $$a_1=-a_2 (r_1+r_2) $$ and $$ a_0 = a_2 (r_1r_2) $$, with which we can for example identify $$ r_1+r_2 = - a_1/a_2 $$ and $$ r_1r_2 = a_0/a_2 $$, which are Vieta's formula's for $$n=2$$.

Alternate proof (mathematical induction)
Vieta's formulas can also be proven by induction as shown below.

Inductive hypothesis:

Let $${P(x)}$$ be polynomial of degree $$n$$, with complex roots $${r_1},{r_2},{\dots},{r_n}$$ and complex coefficients $$a_0,a_1,\dots,a_n$$ where $${ a_n} \neq 0$$. Then the inductive hypothesis is that$${P(x)} = {a_n}{x^n}++{\cdots}++{{a}_{0}} = -{a_n}{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ $$

Base case, $$n = 2 $$ (quadratic):

Let $${a_2},{a_1}$$ be coefficients of the quadratic and $$a_0 $$be the constant term. Similarly, let $${r_1},{r_2}$$ be the roots of the quadratic:$${a_2 x^2}+{a_1 x} + a_0 = {a_2}{(x-r_1)(x-r_2)}$$Expand the right side using distributive property:$${a_2 x^2}+{a_1 x} + a_0 = {a_2}{({x^2}-{r_1x}-{r_2x}+{r_1}{r_2})}$$Collect like terms:$${a_2 x^2}+{a_1 x} + a_0 = {a_2}{({x^2}-{({r_1}+{r_2}){x}}+{r_1}{r_2})}$$Apply distributive property again:$${a_2 x^2}+{a_1 x} + a_0 = {{a_2}{x^2}-{{a_2}({r_1}+{r_2}){x}}+{a_2}{({r_1}{r_2})}}$$The inductive hypothesis has now been proven true for $$n = 2$$.

Induction step:

Assuming the inductive hypothesis holds true for all $$n\geqslant 2$$, it must be true for all $$n+1 $$.$${P(x)} = {a_{n+1}}{x^{n+1}}++{\cdots}++{{a}_{0}}$$By the factor theorem, $${(x-r_{n+1})}$$ can be factored out of $$P(x) $$ leaving a 0 remainder. Note that the roots of the polynomial in the square brackets are $$r_1,r_2,\cdots,r_n$$:$${P(x)} = {(x-r_{n+1})} {[{\frac{{a_ {n+ 1}}{x^ {n+1}}++{\cdots}++{{a}_{0}}}{x- r_{n +1}}}]}$$Factor out $$a_{n+1}$$, the leading coefficient $$P(x)$$, from the polynomial in the square brackets:$${P(x)} ={(a_{n+{1}})}{(x-r_{n+1})} {[{\frac{{x^ {n+1}}+ {\frac{{a_{n}} {x^{n}}}{(a_{n+{1}})}}+{\cdots}+{\frac {a_{1}}{(a_{n+{1}})} {x}}+ } {x- r_{n +1}}}]}$$For simplicity sake, allow the coefficients and constant of polynomial be denoted as $$\zeta$$:$$P(x) = {(a_ {n+1})}{(x-r_ {n+1})}{[{x^n}+{\zeta_{n-1}x^{n-1}}+{\cdots}+{\zeta_0}]}$$Using the inductive hypothesis, the polynomial in the square brackets can be rewritten as:$$P(x) = {(a_ {n+1})} {(x-r_ {n+1})} {[-{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ ]}$$Using distributive property:$$P(x) = {(a_ {n+1})}{({x} {[-{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ ]} {- r_ {n+1}} {[-{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ ]} )}$$After expanding and collecting like terms:$$\begin{align} {P(x)} = -{a_{n+1}}{({r_1}+{r_2}+{\cdots}+{r_n}+{r_{n+1}}){x^{n}}}+{\cdots}+ \\

\end{align}$$The inductive hypothesis holds true for $$n+1$$, therefore it must be true $$\forall n \in \mathbb{N}$$

Conclusion:$${a_ n}{x^n}++{\cdots}++{{a}_{0}} = -{a_n}{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ $$By dividing both sides by $$a_{n}$$, 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, 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."