Vieta's formulas

Any general polynomial of degree n

$$P(x)\; =\; a\_n\; x^n\; +\; a\_\{n-1\}x^\{n-1\}\; +\; \backslash cdots\; +\; a\_1\; x\; +\; a\_0$$

(with the coefficients being real or complex numbers and {{math|a_n ≠ 0}}) has {{math|n}} (not necessarily distinct) complex roots {{math|r₁, r₂, ..., r_n}} by the fundamental theorem of algebra. Vieta's formulas relate the polynomial coefficients to signed sums of products of the roots {{math|r₁, r₂, ..., r_n}} as follows:

{{NumBlk||$$\backslash begin\{cases\}$$

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_2r_3 + r_2r_4+\cdots + r_2r_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}|{{EquationRef|*}}}}

Vieta's formulas can equivalently be written as

for {{math|k {{=}} 1, 2, ..., n}} (the indices {{math|i_k}} are sorted in increasing order to ensure each product of {{math|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 {{EquationNote|*|(*)}} can be solved by Newton's method through an explicit simple iterative formula, the Durand-Kerner method.

Vieta's formulas are frequently used with polynomials with coefficients in any integral domain {{mvar|R}}. Then, the quotients $a\_i/a\_n$ belong to the field of fractions of {{mvar|R}} (and possibly are in {{mvar|R}} itself if $a\_n$ happens to be invertible in {{mvar|R}}) and the roots $r\_i$ are taken in an algebraically closed extension. Typically, {{mvar|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 $a\_n(x-r\_1)(x-r\_2)\backslash dots(x-r\_n)$. 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 $P(x)\backslash neq\; (x-1)(x-3)$. 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 $r\_2=5$.

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

$$r\_1\; +\; r\_2\; =\; -\backslash frac\{b\}\{a\},\; \backslash quad\; r\_1\; r\_2\; =\; \backslash frac\{c\}\{a\}.$$

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

The roots $r\_1,\; r\_2,\; r\_3$ of the cubic polynomial $P(x)\; =\; ax^3\; +\; bx^2\; +\; cx\; +\; d$ satisfy

$$r\_1\; +\; r\_2\; +\; r\_3\; =\; -\backslash frac\{b\}\{a\},\; \backslash quad\; r\_1\; r\_2\; +\; r\_1\; r\_3\; +\; r\_2\; r\_3\; =\; \backslash frac\{c\}\{a\},\; \backslash quad\; r\_1\; r\_2\; r\_3\; =\; -\backslash frac\{d\}\{a\}.$$

Vieta's formulas can be proved by considering the equality

$$a\_n\; x^n\; +\; a\_\{n-1\}x^\{n-1\}\; +\backslash cdots\; +\; a\_1\; x+\; a\_0\; =\; a\_n\; (x-r\_1)\; (x-r\_2)\; \backslash cdots\; (x-r\_n)$$

(which is true since $r\_1,\; r\_2,\; \backslash dots,\; r\_n$ are all the roots of this polynomial), expanding the products in the right-hand side, and equating the coefficients of each power of $x$ between the two members of the equation.

Formally, if one expands $(x-r\_1)\; (x-r\_2)\; \backslash cdots\; (x-r\_n)$ and regroup the terms by their degree in {{tmath|x}}, one gets

:$\backslash sum\_\{k=0\}^n\; (-1)^\{n-k\}x^k\; \backslash left(\backslash sum\_\{\backslash stackrel\{(\backslash forall\; i)\backslash ;\; b\_i\backslash in\backslash \{0,1\backslash \}\}\{b\_1+\backslash cdots+b\_n=n-k\}\}\; r\_1^\{b\_1\}\backslash cdots\; r\_n^\{b\_n\}\backslash right),$

where the inner sum is exactly the {{tmath|k}}th elementary symmetric function

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 $n=2$.

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 $\{r\_1\},\{r\_2\},\{\backslash dots\},\{r\_n\}$ and complex coefficients $a\_0,a\_1,\backslash dots,a\_n$ where $\{\; a\_n\}\; \backslash neq\; 0$. Then the inductive hypothesis is that$$\{P(x)\}\; =\; \{a\_n\}\{x^n\}+\{\{a\_\{n-1\}\}\{x^\{n-1\}\}\}+\{\backslash cdots\}+\{\{a\_\{1\}\}\{x\}\}+\{\{a\}\_\{0\}\}\; =$$

{{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, $n\; =\; 2$

(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 $n\backslash geqslant\; 2$, it must be true for all $n+1$

.$$\{P(x)\}\; =\; \{a\_\{n+1\}\}\{x^\{n+1\}\}+\{\{a\_\{n\}\}\{x^\{n\}\}\}+\{\backslash cdots\}+\{\{a\_\{1\}\}\{x\}\}+\{\{a\}\_\{0\}\}$$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 $r\_1,r\_2,\backslash cdots,r\_n$:$$\{P(x)\}\; =\; \{(x-r\_\{n+1\})\}\; \{[\{\backslash frac\{\{a\_\; \{n+\; 1\}\}\{x^\; \{n+1\}\}+\{\{a\_\{n\}\}\{x^\{n\}\}\}+\{\backslash cdots\}+\{\{a\_\{1\}\}\{x\}\}+\{\{a\}\_\{0\}\}\}\{x-\; r\_\{n\; +1\}\}\}]\}$$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}}+

{{\frac{a_0}{{(a_{n+{1}})}}}}}

{x- r_{n +1}}}]}For simplicity sake, allow the coefficients and constant of polynomial be denoted as $\backslash zeta$:$$P(x)\; =\; \{(a\_\; \{n+1\})\}\{(x-r\_\; \{n+1\})\}\{[\{x^n\}+\{\backslash zeta\_\{n-1\}x^\{n-1\}\}+\{\backslash cdots\}+\{\backslash zeta\_0\}]\}$$Using the inductive hypothesis, the polynomial in the square brackets can be rewritten as:$$P(x)\; =\; \{(a\_\; \{n+1\})\}\; \{(x-r\_\; \{n+1\})\}\; \{[\{\{x^\{n\}\}\}-\{(\{r\_1\}+\{r\_2\}+\{\backslash cdots\}+\{r\_n\})\{x^\{n-1\}\}\}+\{\backslash cdots\}+$$

{{(-1)^{n}}{({r_1}{r_2}{\cdots}{r_n})}}]}Using distributive property:$$P(x)\; =\; \{(a\_\; \{n+1\})\}\{(\{x\}\; \{[\{\{x^\{n\}\}\}-\{(\{r\_1\}+\{r\_2\}+\{\backslash cdots\}+\{r\_n\})\{x^\{n-1\}\}\}+\{\backslash cdots\}+$$

{{(-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:$$\backslash begin\{align\}$$

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

Conclusion:$$\{a\_\; n\}\{x^n\}+\{\{a\_\{n-1\}\}\{x^\{n-1\}\}\}+\{\backslash cdots\}+\{\{a\_\{1\}\}\{x\}\}+\{\{a\}\_\{0\}\}\; =$$

{{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 $a\_\{n\}$, it proves the Vieta's formulas true.

A method similar to Vieta's formula can be found in the work of the 12th century Islamic mathematician Sharaf al-Din al-Tusi. It is plausible that algebraic advancements made by other Islamic mathematician such as Omar Khayyam, al-tusi, and al-Kashi influenced 16th-century algebraists, with Vieta being the most prominent among them.{{Cite journal |last=Ypma |first=Tjalling J. |date=1995 |title=Historical Development of the Newton-Raphson Method |url=https://www.jstor.org/stable/2132904 |journal=SIAM Review |volume=37 |issue=4 |pages=534 |doi=10.1137/1037125 |jstor=2132904 |issn=0036-1445}}{{Cite web |title=François Viète - Biography |url=https://mathshistory.st-andrews.ac.uk/Biographies/Viete/ |access-date=2025-01-07 |website=Maths History |language=en}}

The formulas were derived 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,{{Harv|Funkhouser|1930}} 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.

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