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# Descartes Rule of Signs

As you've studied mathematics, you've probably discovered that it takes time to graph or solve polynomials. If we want to know how many real roots a polynomial $p\left(x\right)$ might have without graphing or solving it, we can use a shortcut called the Descartes Rule of Signs.

Descartes' Rule of Signs states that the possible number of the positive roots of a polynomial is equal to the number of sign changes in the coefficients OR less than the number of sign changes by a multiple of 2. For example, a polynomial with 3 sign changes could have 3 positive roots or 1 positive root.

In this article, we'll explore some of the applications of Descartes' Rule of Signs. Let's get started!

## Applying Descartes' Rule of Signs

Before applying Descartes' Rule of Signs, we must arrange the terms of the polynomial such that the exponents are in descending order (highest exponents first, lowest last). For example, consider the following polynomial:

$f\left(x\right)={x}^{3}+3{x}^{2}-x-{x}^{4}-2$

The exponents are in the wrong order, so let's rewrite the polynomial in descending order:

$f\left(x\right)=-{x}^{4}+{x}^{3}+3{x}^{2}-x-2$

Now, we simply count up how many times we go from a plus sign to a minus sign or vice versa. There are 2: we go from $-{x}^{4}$ to ${x}^{3}$ and then from $3{x}^{2}$ to $-x$ . That means that this polynomial either has 2 positive roots or 0 positive roots. Note that this method won't tell us exactly how many roots a given polynomial has, nor will we learn what those roots are.

## The corollary of Descartes' Rule of Signs

The corollary of Descartes' Rule of Signs states that the possible number of negative roots of a polynomial is equal to the number of sign changes or less than the total number of sign changes by a multiple of 2 after substituting $-x$ for $x$ . The substitution has the effect of negating all of the odd-power terms in the polynomial.

For example, let's say we want to know how many negative roots the following polynomial might have: ${x}^{3}-{x}^{2}-14x+24$

The polynomial's terms are already arranged in descending order by exponent, so we can skip to the substitution step. Since we're swapping $-x$ in for every $x$ , the ${x}^{3}$ becomes $-{x}^{3}$ and the $-14x$ becomes $14x$ . Even powers (including zero) aren't changed. That makes our new polynomial:

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

There is only one sign change: from $-{x}^{2}$ to $14x$ . Therefore, this polynomial has one negative root. We can verify our work by graphing or solving the polynomial in question. You may also be asked for the total number of roots a polynomial might have, meaning that you would use both Descartes' Rule of Signs and its corollary.

## Descartes' Rule of Signs practice questions

a. How many total roots could there be in the following polynomial: $f\left(x\right)={x}^{4}-{x}^{3}+2{x}^{2}-x+27$

First, we double-check to make sure the terms are arranged in descending order (they are). Next, we count the total number of sign changes to determine how many positive roots there may be per Descartes' Rule of Signs. There are 4, meaning that this polynomial has 4, 2, or 0 positive roots.

We're looking for the total number of roots though, which means applying the corollary to Descartes' Rule of Signs as well. Substituting $-x$ for $x$ , we get the following polynomial:

$f\left(x\right)={x}^{4}+{x}^{3}+2{x}^{2}+x+27$

Every coefficient is positive, which means that this polynomial has zero sign changes and therefore zero negative roots. Therefore, our final answer is 4, 2, or 0 total roots.

## Flashcards covering the Descartes Rule of Signs

Precalculus Flashcards

## Get help with Descartes' Rule of Signs

Descartes' Rule of Signs doesn't tell us exactly how many roots a polynomial has or what they are, so it isn't a substitute for solving or graphing polynomials. However, it can save valuable time in certain circumstances. If your student is struggling to remember the steps involved in applying Descartes' Rule of Signs, an experienced tutor could pinpoint why they're having a tough time and address the underlying cause of the issue. Contact Varsity Tutors today for more information.

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