Positive roots edit

The rule states that if the nonzero terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign changes between consecutive (nonzero) coefficients, or is less than it by an even number. A root of multiplicity k is counted as k roots.

In particular, if the number of sign changes is zero or one, the number of positive roots equals the number of sign changes.

Negative roots edit

As a corollary of the rule, the number of negative roots is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than it by an even number. This procedure is equivalent to substituting the negation of the variable for the variable itself. For example, the negative roots of ${\displaystyle ax^{3}+bx^{2}+cx+d}$  are the positive roots of

${\displaystyle a(-x)^{3}+b(-x)^{2}+c(-x)+d=-ax^{3}+bx^{2}-cx+d.}$ 

Thus, applying Descartes' rule of signs to this polynomial gives the maximum number of negative roots of the original polynomial.

Example: cubic polynomial edit

The polynomial

${\displaystyle f(x)=+x^{3}+x^{2}-x-1}$ 

has one sign change between the second and third terms, as the sequence of signs is (+, +, −, −). Therefore, it has exactly one positive root. To find the number of negative roots, change the signs of the coefficients of the terms with odd exponents, i.e., apply Descartes' rule of signs to the polynomial ${\displaystyle f(-x)=-x^{3}+x^{2}+x-1.}$ 

This polynomial has two sign changes, as the sequence of signs is (−, +, +, −), meaning that this second polynomial has two or zero positive roots; thus the original polynomial has two or zero negative roots.

In fact, the factorization of the first polynomial is

${\displaystyle f(x)=(x+1)^{2}(x-1),}$ 

so the roots are −1 (twice) and +1 (once).

The factorization of the second polynomial is

${\displaystyle f(-x)=-(x-1)^{2}(x+1).}$ 

So here, the roots are +1 (twice) and −1 (once), the negation of the roots of the original polynomial.

The following is a rough outline of a proof.^[1] First, some preliminary definitions:

• Write the polynomial ${\displaystyle f(x)}$  as ${\displaystyle \sum _{i=0}^{n}a_{i}x^{b_{i}}}$  where we have integer powers ${\displaystyle 0\leq b_{0}<b_{1}<\cdots <b_{n}}$ , and nonzero coefficients ${\displaystyle a_{i}\neq 0}$ .

• Let ${\displaystyle V(f)}$  be the number of sign changes of the coefficients of ${\displaystyle f}$ , meaning the number of ${\displaystyle k}$  such that ${\displaystyle a_{k}a_{k+1}<0}$ .

• Let ${\displaystyle Z(f)}$  be the number of strictly positive roots (counting multiplicity).

With these, we can formally state Descartes' rule as follows:

If ${\displaystyle b_{0}>0}$ , then we can divide the polynomial by ${\displaystyle x^{b_{0}}}$ , which would not change its number of strictly positive roots. Thus WLOG, let ${\displaystyle b_{0}=0}$ .

Any nth degree polynomial has exactly n roots in the complex plane, if counted according to multiplicity. So if f(x) is a polynomial with real coefficients which does not have a root at 0 (that is a polynomial with a nonzero constant term) then the minimum number of nonreal roots is equal to

${\displaystyle n-(p+q),}$ 

where p denotes the maximum number of positive roots, q denotes the maximum number of negative roots (both of which can be found using Descartes' rule of signs), and n denotes the degree of the equation.

Example: some zero coefficients and nonreal roots edit

The polynomial

${\displaystyle f(x)=x^{3}-1,}$ 

has one sign change; so the maximum number of positive real roots is one. As

${\displaystyle f(-x)=-x^{3}-1,}$ 

has no sign change, the original polynomial has no negative real roots. So the minimum number of nonreal roots is

${\displaystyle 3-(1+0)=2.}$ 

Since nonreal roots of a polynomial with real coefficients must occur in conjugate pairs, it means that x³ − 1 has exactly two nonreal roots and one real root, which is positive.

The subtraction of only multiples of 2 from the maximal number of positive roots occurs because the polynomial may have nonreal roots, which always come in pairs since the rule applies to polynomials whose coefficients are real. Thus if the polynomial is known to have all real roots, this rule allows one to find the exact number of positive and negative roots. Since it is easy to determine the multiplicity of zero as a root, the sign of all roots can be determined in this case.

If the real polynomial P has k real positive roots counted with multiplicity, then for every a > 0 there are at least k changes of sign in the sequence of coefficients of the Taylor series of the function e^axP(x). For sufficiently large a, there are exactly k such changes of sign.^[2][3]

In the 1970s Askold Khovanskii developed the theory of fewnomials that generalises Descartes' rule.^[4] The rule of signs can be thought of as stating that the number of real roots of a polynomial is dependent on the polynomial's complexity, and that this complexity is proportional to the number of monomials it has, not its degree. Khovanskiǐ showed that this holds true not just for polynomials but for algebraic combinations of many transcendental functions, the so-called Pfaffian functions.

• Sturm's theorem – Count of the roots of a polynomial in an interval, without computing them
• Rational root theorem – Relationship between the rational roots of a polynomial and its extreme coefficients
• Geometrical properties of polynomial roots – Geometry of the location of polynomial roots
• Gauss–Lucas theorem – Geometric relation between the roots of a polynomial and those of its derivative

This article incorporates material from Descartes' rule of signs on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

• Descartes' Rule of Signs – Proof of the rule
• Descartes' Rule of Signs – Basic explanation