# Descartes' rule of sign

Descartes' rule of sign is used to determine the number of real zeros of a polynomial function.

It tells us that the number of positive real zeroes in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients. The number of negative real zeroes of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.

We will show how it works with an example.

Example

Determine the number of positive and negative real zeros for the given function (this example is also shown in our video lesson):

$f(x)=x^{5}+4x^{4}-3x^{2}+x-6\\$

Our function is arranged in descending powers of the variable, if it were not we would have to do that as a first step. Second we count the number of changes in sign for the coefficients of f(x).

Here are the coefficients of our variable in f(x):

$1\; \; \;+4\;\;\;-3\;\;\;+1\;\;\;-6$

Our variables goes from positive(1) to positive(4) to negative(-3) to positive(1) to negative(-6).

Between the first two coefficients there are no change in signs but between our second and third we have our first change, then between our third and fourth we have our second change and between our 4th and 5th coefficients we have a third change of coefficients. Descartes´ rule of signs tells us that the we then have exactly 3 real positive zeros or less but an odd number of zeros. Hence our number of positive zeros must then be either 3, or 1.

In order to find the number of negative zeros we find f(-x) and count the number of changes in sign for the coefficients:

$\\ f(-x)=(-x)^{5}+4(-x)^{4}-3(-x)^{2}+(-x)-6=\\ =-x^{5}+4x^{4}-3x^{2}-x-6$

Here we can see that we have two changes of signs, hence we have two negative zeros or less but a even number of zeros..

Totally we have 3 or 1 positive zeros or 2 or 0 negative zeros.