The rule of signs, first described by René Descartes in his work La Géométrie, is a technique for determining the number of positive or negative real roots of a polynomial.

The rule gives us an upper bound number of positive or negative roots of a polynomial. However, it does not tell the exact number of positive or negative roots.

Positive Roots

In order to find the number of positive roots in a polynomial with only one variable, we must first arrange the polynomial by descending variable exponent. For example, $-x^2 + x^3 + x$ would be written $x^3 - x^2 + x$. 

Then, we must count the number of sign differences between consecutive nonzero coefficients. This number, or any number less than it by a multiple of 2, could be the number of positive roots. In the example $x^3 - x^2 + x$, there are two sign changes, after the first and second terms. Thus, there are either two or zero positive roots for this polynomial. 

It is important to note that for polynomials with multiple roots of the same value, each of these roots is counted separately. 

Negative Roots

Finding the negative roots is similar to finding the positive roots. The difference is that you must start by finding the coefficients of odd power (for example, $x^3$ or $x^5$, but not $x^2$ or $x^4$). Once you have located them, multiply each by $-1$. Then the procedure is the same; count the number of sign changes between consecutive nonzero coefficients. This number, or any number less than it by a multiple of 2, could be your number of negative roots. Again it is important to note that multiple roots of the same value should be counted separately.

This can also be done by taking the function, $f(x)$, and substituting the $x$ for $-x$, so that we have the function $f(-x)$. The reason we only bother to change the sign of the odd power coefficients is because if we substitute in $-x$ in an even power, it will just become a positive again. 

For example: $(-x)^3 = (-x)(-x)(-x) = -x^3$ 

but $(-x)^2 = (-x)(-x) = x^2$

We can see that the negative signs cancel out for any even power. By only multiplying the odd powered coefficients by $-1$, we are essentially saving ourselves a step.

Example

Consider the polynomial:

$f(x)=x^3+x^2-x-1$

This function has one sign change between the second and third terms. Therefore it has exactly one positive root. Don't forget that the first term has a sign, which, in this case, is positive. 

Next, we move on to finding the negative roots. Change the exponents of the odd-powered coefficients, remembering to change the sign of the first term. Once you have done this, you have obtained the second polynomial and are ready to find the number of negative roots. This second polynomial is shown below:

$f(-x)=-x^3+x^2+x-1$

This polynomial has two sign changes, after the first and third terms. Therefore, we know that it has at most two negative roots. We know that the number of roots of either sign is the number of sign changes, or a multiple of two less than that. So this polynomial has either $2$ or $0$ negative roots. We can validate this algebraically, as shown below.

First, factor the polynomial:

$f(x)=(x+1)(x+1)(x-1)$.

This simplifies to:

$f(x)=(x+1)^2(x-1)$.

Therefore, the roots are $-1$, $-1$ and $1$.

Complex Roots

A polynomial of $n^{\text{th}}$ degree has exactly $n$ roots. The minimum number of complex roots is equal to:

$n-(p+q)$

where $n$ is the total number of roots in a polynomial, $p$ is the maximum number of positive roots, and $q$ is the maximum number of negative roots.

Example

Consider the polynomial:

$f(x) = x^2+b$

To find the positive roots we count the sign changes. For this example, we will assume that $b>0$. Since there are no sign changes, there are no positive roots $(p=0)$. Now we look for negative roots. Since there are no odd powered coefficients, there are no changes to be made before looking for sign changes; therefore, there are no negative roots $(q = 0)$. Now we apply the complex root equation: $n - (p+ q) = 2 - (0 + 0) = 2$. There are 2 complex roots.