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.
Let be a polynomial with real coefficients. Let be the number of sign changes in the sequence : that is, delete the terms of the sequence that are and let be the number of pairs of consecutive terms in the remaining sequence that have opposite signs. Let be the number of positive roots of (counted with multiplicity). Then is a nonnegative even number.
Let Then the sequence of nonzero coefficients is which changes sign once. So and is nonnegative and even, so has to be There is exactly one positive root.
It is possible to get some information about the negative roots as well: note that and the sequence of coefficients is with two sign changes. So there are either or positive roots of
But the positive roots of correspond to negative roots of So there are or of these. Descartes' rule of signs cannot distinguish between these two possibilities.
In fact, so there is one positive root and two negative roots, and (counted twice because it has multiplicity two).
How many positive and negative roots does have?
By Descartes' rule of signs, the number of sign changes is so there are zero or two positive roots. And 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 or to use the intermediate value theorem. In particular, and so there is a root between and 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.)
Suppose that is a polynomial of degree with real coefficients which is expressible as the sum of at most monomials. Suppose also that Show that has at least one non-real root.
The number of sign changes in the coefficients of is at most as is the number of sign changes in So, by Descartes' rule of signs, the number of positive roots plus the number of negative roots is at most
If is even, this number is and if is odd it is In both cases it is less than
The fundamental theorem of algebra says that has exactly complex roots, with multiplicity, and the above argument shows that there are fewer than real roots, so has at least one non-real root.
A specific example to illustrate the above: Suppose where are any real numbers and Then has at least one non-real root.
There are two parts. The first is to show that is even; the second is to show that it is nonnegative.
Without loss of generality, assume that has a positive leading coefficient. Now suppose the rightmost coefficient, is positive as well. Then is even, since the signs of the coefficients start and end with and the graph of crosses the -axis an even number of times as well (since it starts and ends above the -axis). So is even. If is negative, similar arguments show that both and are odd, so is even.
To show that induct on The base case is clear. Suppose that all polynomials of degree satisfy and take a polynomial of degree Consider the polynomial of degree The number of sign changes of is either or depending on whether there is a sign change at the rightmost coefficient of But the number of positive zeroes of is at least by Rolle's theorem, since between any two zeroes of there is a zero of
Since has degree the inductive hypothesis implies that but then
so but is even, so it must be nonnegative. This proves the theorem by induction.