# Descartes' Rule of Signs

**Descartes' rule of signs** is a criterion which gives an upper bound on the number of positive or negative real roots of a polynomial with real coefficients. The bound is based on the number of sign changes in the sequence of coefficients of the polynomial.

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## Statement of Descartes' Rule of Signs

Let \( f(x) = a_nx^n + a_{n-1}x^{n-1}+ \cdots+a_0\) be a polynomial with real coefficients. Let \( s\) be the number of sign changes in the sequence \( a_n,a_{n-1},\ldots,a_0\): that is, delete the terms of the sequence that are \( 0,\) and let \( s \) be the number of pairs of consecutive terms in the remaining sequence that have opposite signs. Let \( p \) be the number of positive roots of \( f(x) \) (counted with multiplicity). Then \( s-p \) is a nonnegative even number.

Let \( f(x) = x^3-3x-2.\) Then the sequence of nonzero coefficients is \(1,-3,-2,\) which changes sign once. So \( s=1,\) and \( s-p\) is nonnegative and even, so \( p\) has to be \( 1.\) There is exactly one positive root.

It is possible to get some information about the negative roots as well: note that \( f(-x) = (-x)^3-3(-x)-2 = -x^3+3x-2,\) and the sequence of coefficients is \( -1,3,-2,\) with two sign changes. So there are either \( 0\) or \(2\) positive roots of \( f(-x).\)

But the positive roots of \( f(-x)\) correspond to negative roots of \( f(x).\) So there are \( 0\) or \( 2\) of these. Descartes' rule of signs cannot distinguish between these two possibilities.

In fact, \( x^3-3x-2 = (x+1)^2(x-2),\) so there is one positive root \( 2,\) and two negative roots, \( -1\) and \(-1\) (counted twice because it has multiplicity two).

## Applications of Descartes' Rule of Signs

How many positive and negative roots does \( x^3-3x^2+1\) have?

By Descartes' rule of signs, the number of sign changes is \( 2,\) so there are zero or two positive roots. And \( f(-x) = -x^3-3x^2+1\) has one sign change, so there is exactly one negative root.

To decide whether there are zero or two positive roots, it is a good idea to look at the graph of \( y=f(x),\) or to use the intermediate value theorem. In particular, \( f(0) = 1 \) and \( f(1) = -1,\) so there is a root between \(0\) and \( 1.\) So there can't be zero positive roots, so there must be two. (There is also a root between 2 and 3, by a similar IVT argument.) \(_\square\)

Suppose that \( f(x) \) is a polynomial of degree \( n \) with real coefficients which is expressible as the sum of at most \( \big\lceil \frac n2 \big\rceil\) monomials. Suppose also that \( f(0) \ne 0.\) Show that \( f(x) \) has at least one non-real root.

The number of sign changes in the coefficients of \( f(x) \) is at most \( \big\lceil \frac n2 \big\rceil-1,\) as is the number of sign changes in \( f(-x).\) So, by Descartes' rule of signs, the number of positive roots plus the number of negative roots is at most

\[2 \left\lceil \frac n2 \right\rceil - 2.\]

If \( n\) is even, this number is \( n-2,\) and if \(n\) is odd it is \( n-1.\) In both cases it is less than \( n.\)

The fundamental theorem of algebra says that \( f\) has exactly \(n\) complex roots, with multiplicity, and the above argument shows that there are fewer than \(n\) real roots, so \( f\) has at least one non-real root. \(_\square\)

A specific example to illustrate the above: Suppose \( f(x) = ax^5+bx^2+1,\) where \(a,b\) are any real numbers and \( a\ne 0.\) Then \(f(x) \) has at least one non-real root.

## Proof of Descartes' Rule of Signs

There are two parts. The first is to show that \( s-p \) is even; the second is to show that it is nonnegative.

Without loss of generality, assume that \( f(x) \) has a positive leading coefficient. Now suppose the rightmost coefficient, \( f(0),\) is positive as well. Then \( s\) is even, since the signs of the coefficients start and end with \( +,\) and the graph of \(y=f(x) \) crosses the \( x\)-axis an even number of times as well (since it starts and ends above the \(x\)-axis). So \( s-p\) is even. If \( f(0)\) is negative, similar arguments show that both \( s\) and \( p\) are odd, so \( s-p\) is even.

To show that \( s\ge p,\) induct on \(n.\) The base case \( n=1 \) is clear. Suppose that all polynomials of degree \( n-1\) satisfy \( s\ge p,\) and take a polynomial \(f(x)\) of degree \(n.\) Consider the polynomial \( f'(x)\) of degree \( n-1.\) The number of sign changes \( s'\) of \( f'(x) \) is either \( s\) or \( s-1,\) depending on whether there is a sign change at the rightmost coefficient of \(f.\) But the number of positive zeroes \( p'\) of \( f'(x) \) is at least \( p-1\) by Rolle's theorem, since between any two zeroes of \(f\) there is a zero of \( f'.\)

Since \( f'(x) \) has degree \( n-1,\) the inductive hypothesis implies that \( s'\ge p',\) but then

\[s \ge s' \ge p' \ge p-1,\]

so \( s-p \ge -1,\) but \( s-p \) is even, so it must be nonnegative. This proves the theorem by induction.

**Cite as:**Descartes' Rule of Signs.

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