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 want to 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:
• Written in descending powers of x
• Has only real coefficients
• Has a nonzero constant term

### Descartes' Rule of Signs for Positive 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 we 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 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 we decrease this number by 2 until we get to zero or can't subtract away 2 without going negative
Let's look at an example.
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
+4No0
-2Yes1
+2Yes2
-1Yes3
-2No3
+1Yes4
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
-4No0
-2No0
-2No0
-1No0
+2Yes1
+1No1
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.

#### 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: 0

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