Lesson Objectives
• Learn how to find the possible number of positive real zeros
• Learn how to find the possible number of negative real zeros

## How to Use Descartes' Rule of Signs

In this lesson, we will learn about Descartes' Rule of Signs. When trying to find the zeros for a polynomial function, Descartes' Rule of Signs can be helpful. It tells us the possible number of positive and negative real zeros for a polynomial function that meets the following conditions: $$f(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0$$
• Written in descending powers of x
• Has only real coefficients
• Has a nonzero constant term

### Variation in Sign

A variation in sign or a sign change is a change from positive to negative or negative to positive in consecutive terms of our polynomial function when written in descending powers of x. Any missing terms will be ignored and counted as no sign change. $$f(x) = 2x^6 - 3x^4 - x^3 + x^2 - 1$$ In the function above, we have 3 sign changes. Starting with +2 as the coefficient for x6, when we move right, we end up with -3x4, which has a coefficient of -3. This would give us the first sign change as we went from +2 to -3. Next, we move on to -x3, which has a coefficient of -1. Since we are going from -3 to -1, there is no sign change here, both are negative. Next, we move on to x2, which has a coefficient of +1. This would give us the second sign change as we went from -1 to +1. Finally, we have our constant term, which is -1. This would give us the third sign change as we went from +1 to -1. The results are summarized in a table below:
Coefficient Change Count
+2 No 0
-3 Yes 1
-1 No 1
+1 Yes 2
-1 Yes 3

### Descartes' Rule of Signs for Positive Real Zeros

• Count the number of sign changes that occur in the coefficients of the function
• The number of positive real zeros is equal to the number found above or is less than that number by some positive even integer
• Summary: find the number of sign changes, then decrease this number by 2 until we get to zero or can't subtract away 2 without going negative

### Descartes' Rule of Signs for Negative Real Zeros

• Plug in (-x) for each occurrence of x, in other words, find f(-x)
• Count the number of sign changes that occur in the coefficients of the function
• The number of negative real zeros is equal to the number found above or is less than that number by some positive even integer
• Summary: find f(-x), then find the number of sign changes, then decrease this number by 2 until we get to zero or can't subtract away 2 without going negative
Let's look at some examples.
Example #1: Determine the possible number of positive and negative real zeros. $$f(x)=4x^5 - 2x^4 + 2x^3 - x^2 - 2x + 1$$ We will begin by thinking about the possible number of positive real zeros.
How many sign changes? Let's think about the coefficients:
Coefficient Change Count
+4 No 0
-2 Yes 1
+2 Yes 2
-1 Yes 3
-2 No 3
+1 Yes 4
Since there are 4 sign changes, we can conclude that there are either 4, 2, or 0 positive real zeros. Let's think about the possible number of negative real zeros: $$f(-x)=-4x^5 - 2x^4 - 2x^3 - x^2 + 2x + 1$$ How many sign changes? Let's think about the coefficients:
Coefficient Change Count
-4 No 0
-2 No 0
-2 No 0
-1 No 0
+2 Yes 1
+1 No 1
There is only 1 sign change in f(-x), which tells us there will be 1 negative real zero. We can't subtract 2 from 1 without going negative, so this is our final answer.
Example #2: Determine the possible number of positive and negative real zeros. $$f(x)=9x^5 - 6x^4 + 66x^3 - 44x^2 + 72x - 48$$ We will begin by thinking about the possible number of positive real zeros.
How many sign changes? Let's think about the coefficients:
Coefficient Change Count
+9 No 0
-6 Yes 1
+66 Yes 2
-44 Yes 3
+72 Yes 4
-48 Yes 5
Since there are 5 sign changes, we can conclude that there are either 5, 3, or 1 positive real zeros. Let's think about the possible number of negative real zeros: $$f(-x)=-9x^5 - 6x^4 - 66x^3 - 44x^2 - 72x - 48$$ How many sign changes? Let's think about the coefficients:
Coefficient Change Count
-9 No 0
-6 No 0
-66 No 0
-44 No 0
-72 No 0
-48 No 0
There are no sign changes in f(-x), which tells us we will have 0 negative real zeros.

### Constructing a Table of All Possible Zeros

When we combine the Fundamental Theorem of Algebra, the Conjugate Zeros Theorem, and Descartes' rule of signs, we can come up with a table that describes all of the possible outcomes. Let's look at an example.
Example #3: Create a table that gives all the possible positive real zeros, negative real zeros, and non-real complex zeros. $$f(x) = 25x^6 - 25x^4 - 16x^2 + 16$$ Here, we have 2 sign changes, which means we have either 2 positive real zeros or 0 positive real zeros. $$f(-x) = 25x^6 - 25x^4 - 16x^2 + 16$$ Once again, we have 2 sign changes, which means we have either 2 negative real zeros or 0 negative real zeros.
From the Fundamental Theorem of Algebra, we know that a polynomial function of degree 6 will have exactly 6 zeros when we include multiplicities. When we set up our table, we will need the total in each case to be 6. Additionally, from the Conjugate Zeros Theorem, we know that non-real complex zeros will occur in conjugate pairs. Putting all this information together allows us to create a table for all of the possibilities. We will refer to the number of possible positive real zeros as "Positive", the number of possible negative real zeros as "Negative", and the number of possible non-real complex zeros as "Imaginary" in our table below.
Positive Negative Imaginary Total
2 2 2 6
2 0 4 6
0 2 4 6
0 0 6 6

#### Skills Check:

Example #1

State the possible number of positive and negative real zeros. $$f(x)=27x^6 - 35x^3 + 8$$

A
+ zeros: 4, 2, or 0, - zeros: 3 or 1
B
+ zeros: 2 or 0, - zeros: 0
C
+ zeros: 0, - zeros: 1 or 3
D
+ zeros: 1 or 3, - zeros: 0
E
+ zeros: 2 or 0, - zeros: 1

Example #2

State the possible number of positive and negative real zeros. $$f(x)=5x^6 - 3x^4 - 80x^2 + 48$$

A
+ zeros: 0, - zeros: 3 or 1
B
+ zeros: 2 or 0, - zeros: 3 or 1
C
+ zeros: 4, 2, or 0, - zeros: 2 or 0
D
+ zeros: 2 or 0, - zeros: 2 or 0
E
+ zeros: 4, 2, or 0, - zeros: 3 or 1

Example #3

State the minimum number of possible non-real complex zeros. $$f(x)=3x^6 - 4x^4 - 27x^2 + 36$$

A
6
B
4
C
2
D
0
E
12