1. WARM UP Find all real zeros of the functions
1. f(x) = x3 – 3x2 – 6x + 8.

2. Homework Questions

3. Homework Questions

4. Homework Questions

5. Fundamental Theorem of Algebra Descartes’ Rule of Signs
EQ: How can you determine the possible number of positive, negative, and imaginary zeros of a polynomial function?
Assessment: Students will write a summary about using Descartes Rule of Signs to determine all the zeros of a polynomial function.

6. Key Concept:

7. Use Descartes’ Rule of Signs
STEP 1 – Determine Positive Zeros by counting the number of sign changes in
the function.
f (x) = x6 – 2x5 + 3x4 – 10x3 – 6x2 – 8x – 8.
The coefficients in f (x) have 3 sign changes, so f has 3 or 1 positive real zero(s).
STEP 2 – Determine Negative Zeros by changing f(x) to f(-x), then counting
the number of sign changes in the function.
f (– x) = (– x)6 – 2(– x)5 + 3(– x)4 – 10(– x)3 – 6(– x)2 – 8(– x) – 8
= x6 + 2x5 + 3x4 + 10x3 – 6x2 + 8x – 8

8. Use Descartes’ Rule of Signs
The coefficients in f (– x) have 3 sign changes, so f has 3 or 1 negative real zero(s) .
The possible numbers of zeros for f are summarized in the table below.

9. The coefficients in g(– x) have no sign changes.
Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.
g(x) = 2x4 – 8x3 + 6x2 – 3x + 1
g(x) = 2x4 – 8x3 + 6x2 – 3x + 1
The coefficients in g(x) have 4 sign changes, so f has 4 positive real 0’(s).
g(– x) = 2(– x)4 – 8(– x)3 + 6(– x)2 + 1
= 2x4 + 8x + 6x2 + 1
The coefficients in g(– x) have no sign changes.
The possible numbers of zeros for f are summarized in the table below.

10. You Try It
Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.
f (x) = x3 + 2x – 11
f (x) = x3 + 2x – 11
f (– x) = (– x)3 + 2(– x) – 11
= – x3 – 2x – 11
The possible numbers of zeros for f are summarized in the table below.

11. HOMEWORK Textbook – Pg. 384 # 34-39
Write your Summary to the Essential Question