Descartes' Rule of Signs

Example 1:

Find the number of the positive roots of the polynomial.

$x^3+3x^2-x-x^4-2$

Arrange the terms of the polynomial in the descending order of exponents:

$-x^4+x^3+3x^2-x-2$

Count the number of sign changes:

$\stackrel{1}{\stackrel{⎴}{-x^4+}}{x}^3\stackrel{2}{\stackrel{⎴}{+3x^2-}}x-2$

There are $2$ sign changes in the polynomial, so the possible number of positive roots of the polynomial is $2$ or $0$ .

Example 2:

Find the possible number of real roots of the polynomial and verify.

$x^3-x^2-14x+24$

The terms of the polynomial are already in the descending order of exponents.

Count the number of sign changes:

$\stackrel{1}{\stackrel{⎴}{+x^3-}}{x}^2\stackrel{2}{\stackrel{⎴}{-14x+}}24$

There are $2$ sign changes in the polynomial and the possible number of positive roots of the polynomial is $2$ or $0$ .

Let the given polynomial be $f\left(x\right)$ and substitute $-x$ for $x$ in the polynomial and simplify:

$\begin{array}{l}f\left(-x\right)={\left(-x\right)}^3-{\left(-x\right)}^2-14\left(-x\right)+24\\ =-x^3-x^2+14x+24\end{array}$

Count the number of sign changes:

$-x^3\stackrel{1}{\stackrel{⎴}{-x^2+}}14x+24$

There is $1$ sign change in the second polynomial. So, from the corollary of the Descartes' Rule of Signs, the possible number of negative roots of the original polynomial is $1$ .

The polynomial can be rewritten as: $\left(x-2\right)\left(x-3\right)\left(x+4\right)$

We can verify that there are $2$ positive roots and $1$ negative root of the given polynomial.

Please note that the repeated roots of a polynomial are counted separately.

For example, the polynomial

${\left(x-2\right)}^2$ , which can be written as $x^2-2x+4$ , has $2$ sign changes. Therefore, the polynomial has $2$ positive roots.