# Descartesâ€™ Rule of Signs

The purpose of the Descartes’ Rule of Signs is to provide an insight on how many real roots a polynomial [latex]P\left( x \right)[/latex] may have. We are interested in two kinds of real roots, namely **positive** and **negative** real roots. The rule is actually simple.

Here is the Descartes’ Rule of Signs in a nutshell.

## Breakdown or Explanation of the Descartes’ Rule of Signs

Suppose [latex]P\left( x \right)[/latex] is a polynomial where the exponents are arranged from highest to lowest, with real coefficients excluding zero, and contains a nonzero constant term.

The number of **positive** real roots is either

- equal to the number of sign changes in [latex]P\left( x \right)[/latex]

**or**, less than the number of sign changes in [latex]P\left( x \right)[/latex] by some multiple of [latex]2[/latex].

The number of **negative** real roots is either

- equal to the number of sign changes in [latex]P\left( { – x} \right)[/latex]

**or**, less than the number of sign changes in [latex]P\left( { – x} \right)[/latex] by some multiple of [latex]2[/latex].

In summary, if [latex]n[/latex] is the number of sign changes in either [latex]P\left( x \right)[/latex] or [latex]P\left( { – x} \right)[/latex], then the number of positive or negative roots may equal to [latex]n[/latex], [latex]n-2[/latex], [latex]n-4[/latex], [latex]n-6[/latex], etc.

Note that we start with the number of sign changes, “[latex]n[/latex]”, then we keep subtracting it by some multiple of [latex]2[/latex] (positive even integers) such as [latex]2[/latex], [latex]4[/latex], [latex]6[/latex], etc.

We stop subtracting until such time when the difference becomes [latex]0[/latex] or [latex]1[/latex]. That is it!

**Quick examples for both cases:**

- Let [latex]P\left( x \right)[/latex] has [latex]n=7[/latex] number of sign changes, the possible number of positive real roots will be

[latex]7[/latex], [latex]5[/latex], [latex]3[/latex] or [latex]1[/latex]

- Let [latex]P\left( x \right)[/latex] has [latex]n = 6[/latex] number of sign changes, the possible number of negative real roots will be

[latex]6[/latex], [latex]4[/latex], [latex]2[/latex] or [latex]0[/latex]

### Examples of Descartes’ Rule of Signs

Let’s take a look at a few examples to see this rule in action!

**Example 1**: Find the number of real roots of the polynomial below using Descartes’ Rule of Signs.

Start by clearly marking off the sign of each term in the polynomial. I will use the color red for positive symbol (+), and **black** for negative symbol (**âˆ’**). This would allow us to easily keep track of the change in sign.

It is considered a **sign change** if the two signs of adjacent coefficients switch (or alternate). For instance, it can go from positive to negative, or negative to positive.

**For the positive real roots**:

Use the given function itself because the “[latex]x[/latex]” inside the parenthesis, [latex]P\left( x \right)[/latex], is positive.

There are two sign changes as shown by the arrows. Since [latex]n = 2[/latex], therefore there are [latex]2[/latex] or [latex]0[/latex] positive real roots.

**For the negative real roots**:

Use the modified version of the function, [latex]P\left( { – x} \right)[/latex], where the “[latex]x[/latex]” inside the parenthesis is negative.

Before we count the sign change, we will need some side calculation. Substitute “[latex]-x[/latex]” in [latex]P\left( x \right)[/latex] to get [latex]P\left( { – x} \right)[/latex]. Here we go…

Now, let’s do the counting…

There are three sign changes as pointed out by the arrows. Since [latex]n = 3[/latex], therefore there are [latex]3[/latex] or [latex]1[/latex] negative real roots.

For our final answer, we say, there are [latex]2[/latex] or [latex]0[/latex] positive real roots, and [latex]3[/latex] or [latex]1[/latex] negative real roots.

Here is the graph of the polynomial showing that indeed our “guess” is spot on! In fact, it has two (2) positive roots, and three (3) negative roots.

**Example 2**: Find the number of real roots of the polynomial below using Descartes’ Rule of Signs.

Before we start on this problem, I must caution you **not** to treat this like a synthetic division problem where we place zeroes on the missing powers of [latex]x[/latex]’s. As long as the polynomial is arranged with a decreasing number of exponents, that is good enough.

**For the positive real roots**:

Use the given polynomial and count the number of sign change.

There are three sign changes for [latex]P\left( x \right)[/latex], that means there can be [latex]3[/latex] **or **[latex]1[/latex] positive real roots.

**For the negative real roots**:

Evaluate [latex]-x[/latex] into [latex]P\left( x \right)[/latex] to get [latex]P\left( { – x} \right)[/latex] then count the sign change.

There is only one sign change for [latex]P\left( { – x} \right)[/latex], that means there is **exactly 1** negative real root.

For our final answer, there are [latex]3[/latex] or [latex]1[/latex] positive real roots, and exactly [latex]1[/latex] negative real root.

**Example 3**: Find the number of real roots (positive and/or negative) of the polynomial below.

**To find the positive roots**:

Observe that all terms in the polynomial are all positive.

Since there is no sign change in [latex]P\left( x \right)[/latex], this implies that the polynomial has **NO** positive real roots.

**To find the negative roots**:

Solve for [latex]P\left( { – x} \right)[/latex] then count the variation in signs.

Because we have seven (7) sign changes in [latex]P\left( { – x} \right)[/latex], there are [latex]7[/latex], [latex]5[/latex], [latex]3[/latex], or [latex]1[/latex] negative real roots.

For our final answer, there are no positive real roots, and there are [latex]7[/latex], [latex]5[/latex], [latex]3[/latex], or [latex]1[/latex] negative real roots.

**Example 4**: Find the number of real roots of the polynomial (positive and/or negative) below.

**To find the positive roots**:

Count the number of alternating signs in [latex]P\left( x \right)[/latex].

We have six sign changes here which imply that there are [latex]6[/latex], [latex]4[/latex], [latex]2[/latex], or [latex]0[/latex] positive real roots.

**To find the negative roots**:

Solve for [latex]P\left( { – x} \right)[/latex] first then count the variation in signs.

No sign change in [latex]P\left( { – x} \right)[/latex] that means there are no negative real solutions.

Therefore, there are [latex]6[/latex], [latex]4[/latex], [latex]2[/latex], or [latex]0[/latex] positive real roots, and there are no negative real roots.