# Descartes' Rule of Signs

Descartes' Rule of Signs helps to identify the possible number of
real roots
of a polynomial
$p\left(x\right)$
without actually graphing or solving it. Please note that this rule does not give the
*
exact number of roots
*
of the polynomial or
*
identify the roots
*
of the polynomial.

The rule states that the possible number of the positive roots of a polynomial is equal to the number of sign changes in the coefficients of the terms or less than the sign changes by a multiple of $2$ .

For example, if there are $3$ sign changes in the coefficients of the terms of polynomial, then the possible number of positive roots of the polynomial is $3$ or $1$ .

[Before applying the Descartes' Rule of Signs, make sure to arrange the terms of the polynomial in descending order of exponents .]

**
Example 1:
**

Find the number of the positive roots of the polynomial.

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

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

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

Count the number of sign changes:

$\stackrel{{1}}{\stackrel{{\u23b4}}{-{x}^{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+}}{x}^{3}\stackrel{{2}}{\stackrel{{\u23b4}}{+3{x}^{2}-}}x-2$

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

### Corollary of Descartes' Rule of Signs:

First rewrite the given polynomial by substituting $-x$ for $x$ . This is same as negating the coefficients of the odd-power terms.

The corollary rule states that the possible number of the negative roots of the original polynomial is equal to the number of sign changes (in the coefficients of the terms after negating the odd-power terms) or less than the sign changes by a multiple of $2$ .

**
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:

$\stackrel{{1}}{\stackrel{{\u23b4}}{+{x}^{3}-}}{x}^{2}\stackrel{{2}}{\stackrel{{\u23b4}}{-14x+}}24$

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:

$\begin{array}{l}f\left(-x\right)={\left(-x\right)}^{3}-{\left(-x\right)}^{2}-14\left(-x\right)+24\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-{x}^{3}-{x}^{2}+14x+24\end{array}$

Count the number of sign changes:

$-{x}^{3}\stackrel{{1}}{\stackrel{{\u23b4}}{-{x}^{2}+}}14x+24$

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.