1. Name:__________ warm-up 5-7
Use synthetic substitution to find f(2) for f(r) = 3r4 + 7r2 – 12r + 23.
Use synthetic substitution to find f(6) for f(c) = 2c3 + 19c2 + 2.
Given a polynomial and one of its factors, find the remaining factors of the polynomial. k4 + 7k3 + 9k2 – 7k – 10; k + 2
Given a polynomial and one of its factors, find the remaining factors of the polynomial. 6p3 + 11p2 – 14p – 24; p + 2

2. The function f(x) = x3 – 6x2 – x + 30 can be used to describe the relative stability of a small boat carrying x passengers, where f(x) = 0 indicates that the boat is extremely unstable. With three passengers, the boat tends to capsize. What other passenger loads could cause the boat to capsize
What value of k would give a remainder of 6 when x2 + kx + 18 is divided by x + 4?

3. 5-7 Roots and Zeroes x 4 – 29x 2 + 100 = 0 x 6 – 35x 3 + 216 = 0.

4. Details of the Day Activities: EQ:
How do polynomials functions model real world problems and their solutions?
I will be able to…
Determine the number and type of roots for a polynomial equation.
Activities:
Warm-up
Review homework
Notes:
Marking Period Exam – November 12
. Find the zeros of a polynomial function.

5. A Quick Review
Use synthetic substitution to find f(2) for f(r) = 3r4 + 7r2 – 12r + 23.
Use synthetic substitution to find f(6) for f(c) = 2c3 + 19c2 + 2.
Given a polynomial and one of its factors, find the remaining factors of the polynomial. k4 + 7k3 + 9k2 – 7k – 10; k + 2
Given a polynomial and one of its factors, find the remaining factors of the polynomial. 6p3 + 11p2 – 14p – 24; p + 2

6. A Quick Review
The function f(x) = x3 – 6x2 – x + 30 can be used to describe the relative stability of a small boat carrying x passengers, where f(x) = 0 indicates that the boat is extremely unstable. With three passengers, the boat tends to capsize. What other passenger loads could cause the boat to capsize
What value of k would give a remainder of 6 when x2 + kx + 18 is divided by x + 4?

7. Notes and examples

8. Notes and examples
. Solve x2 + 2x – 48 = 0. State the number and type of roots.
. Solve y4 – 256 = 0. State the number and types of roots.

9. Notes and examples
Solve x2 – x – 12 = 0. State the number and type of roots.
Solve a4 – 81 = 0. State the number and type of roots.

10. Notes and examples

11. Notes and examples yes – to + yes + to – no – to – no – to –
State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x) = –x6 + 4x3 – 2x2 – x – 1.
Since p(x) has degree 6, it has 6 zeros. However, some of them may be imaginary. Use Descartes’ Rule of Signs to determine the number and type of real zeros. Count the number of changes in sign for the coefficients of p(x).
p(x) = –x6 + 4x3 – 2x2 – x – 1
Since there are two sign changes, there are 2 or 0 negative real zeros. Make a chart of possible combinations.
yes
– to +
yes
+ to – no
– to – no
– to –
There are 2 or 0 positive real zeros, 2 or 0 negative real zeros, and 6, 4, or 2 imaginary zeros.

12. Notes and examples
State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x) = x4 – x3 + x2 + x + 3.
A. positive: 2 or 0; negative: 3 or 1; imaginary: 1, 3, or 5
B. positive: none; negative: none; imaginary: 6
C. positive: 2 or 0; negative: 0; imaginary: 6 or 4
D. positive: 2 or 0; negative: 2 or 0; imaginary: 6, 4, or 2

13. Notes and examples f(–x) = –x3 – x2 – 2x + 4 yes yes no no no yes
Find all of the zeros of f(x) = x3 – x2 + 2x + 4.
Since f(x) has degree of 3, the function has three zeros. To determine the possible number and type of real zeros, examine the number of sign changes in f(x) and f(–x).
f(x) = x3 – x2 + 2x + 4
yes
yes no
f(–x) = –x3 – x2 – 2x + 4 no no
yes

14. Notes and examples
The function has 2 or 0 positive real zeros and exactly 1 negative real zero. Thus, this function has either 2 positive real zeros and 1 negative real zero or 2 imaginary zeros and 1 negative real zero.
To find the zeros, list some possibilities and eliminate those that are not zeros. Use synthetic substitution to find f(a) for several values of a.
From the table, we can see that one zero occurs at x = –1. Since the depressed polynomial, x2 – 2x + 4, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation x2 – 2x + 4 = 0.
Each row in the table shows the coefficients of the depressed polynomial and the remainder

15. Notes and examples

16. Notes and examples
Answer: Thus, this function has one real zero at –1 and two imaginary zeros at The graph of the function verifies that there is only one real zero.

17. Notes and examples
What are all the zeros of f(x) = x3 – 3x2 – 2x + 4? A.
B C. D.

18. Notes and examples

19. Notes and examples

20. Notes and examples

21. Notes and examples

22. Notes and examples