# Descartes' rule of signs

In mathematics, Descartes' rule of signs, first described by René Descartes in his work La Géométrie, is a technique for determining an upper bound on the number of positive or negative real roots of a polynomial. It is not a complete criterion, because it does not provide the exact number of positive or negative roots.

The rule is applied by counting the number of sign changes in the sequence formed by the polynomial's coefficients. If a coefficient is zero, that term is simply omitted from the sequence.

## Descartes' rule of signs

### Positive roots

The rule states that if the 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 differences between consecutive nonzero coefficients, or is less than it by an even number. Multiple roots of the same value are counted separately.

### Negative roots

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, to find the number of negative roots of ${\displaystyle f(x)=ax^{3}+bx^{2}+cx+d}$, we equivalently ask how many positive roots there are for ${\displaystyle -x}$ in ${\displaystyle f(-x)=a(-x)^{3}+b(-x)^{2}+c(-x)+d=-ax^{3}+bx^{2}-cx+d\equiv g(x).}$ Using Descartes' rule of signs on ${\displaystyle g(x)}$ gives the number of positive roots ${\displaystyle x_{i}}$ of g, and since ${\displaystyle g(x)=f(-x)}$ it gives the number of positive roots ${\displaystyle (-x_{i})}$ of f, which is the same as the number of negative roots ${\displaystyle x_{i}}$ of f.

### Example: real roots

The polynomial

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

has one sign change between the second and third terms (the sequence of pairs of successive signs is ++, +−, −−). Therefore it has exactly one positive root. Note that the leading sign needs to be considered although in this particular example it does not affect the answer. 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 ${\displaystyle f(-x)}$, to obtain a second polynomial

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

This polynomial has two sign changes (the sequence of pairs of successive 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.

The factorization of the second polynomial is

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

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

## Nonreal roots

Any nth degree polynomial has exactly n roots in the complex plane, if counted according to multiplicity. So if f(x) is a polynomial which does not have a root at 0 (which can be determined by inspection) 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: zero coefficients, nonreal roots

The polynomial

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

has one sign change, so the maximum number of positive real roots is 1. From

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

we can tell that the 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, we can see that x3 − 1 has exactly 2 nonreal roots and 1 real (and positive) root.

## Special case

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.

## Generalizations

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 eaxP(x). For a sufficiently large, there are exactly k such changes of sign.[1][2]

In the 1970s Askold Georgevich Khovanskiǐ developed the theory of fewnomials that generalises Descartes' rule.[3] 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.