Descartes' Rule of Signs Lesson

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
+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.

---

Skills Check: