1. Warm-up:
Find f(2) for
HW: pg. 248(2 – 14even, 18, 20, 26, 31, 46, 48)

2. 55) (x – 2)(x + 3)(x – 1);zeros 2, -3, 1
HW: Page 239 (7-12 by any method, 24, 27, 30, 33, 36, 39) page 240 (49, 51, 55, 57, 59)
7) 2x + 4
8) 5x + 3
9) x2 – 3x + 1
10) 6x2 – 12x + 9
11) x3 + 3x2 – 1
12) x + 4
24) 5x2 + 3x – 2
27) –x2 + 10x – 25
30) 5x2 – 10x + 26 –
x2 – 8x + 64
36) –3x3 + 6x2 – 12x + 24 –
39) 4x2 + 14x – 30
49) a.1 b.4 c.4 d.1954
51) a.97 b.-5/3 c.17 d.-199
55) (x – 2)(x + 3)(x – 1);zeros 2, -3, 1
57) (2x – 1)(x – 5)(x – 2); zeros ½, 5, 2
59)

3. The Real Zeros of a Polynomial Function
Objective:
Use Descartes Rule of Signs
Use the Rational Zeros Test
Find bounds for real zeros of polynomial functions
Factor polynomials completely
Find real zeros of a polynomial

4. Our goal in this section is to learn how we can factor higher degree polynomials such as:
We are not given any factors of the function.
We can use synthetic division to find a factor by trial and error.
We shall learn a few techniques to hasten the trial and error method.
The first of these is called Descartes Rule of Signs named after a French mathematician that worked in the 1600’s.
Rene Descartes

5. Descartes’ Rule of Signs
Let f denote a polynomial function written in standard form.
The number of positive real zeros of f either equals the number of sign changes of f (x) or else equals that number less an even integer.
The number of negative real zeros of f either equals the number of sign changes of f (-x) or else equals that number less an even integer. 1 2
starts Pos changes Neg. changes Pos.
2 sign changes  could be 2 or 0 positive real zeros to the polynomial.

6. Descartes’Rule of Signs
Let f denote a polynomial function written in standard form.
The number of positive real zeros of f either equals the number of sign changes of f (x) or else equals that number less an even integer.
The number of negative real zeros of f either equals the number of sign changes of f (-x) or else equals that number less an even integer. 1 2
starts Pos. changes Neg. changes Pos.
2 sign changes could be 2 or 0 negative real zeros to the polynomial.
simplify f(-x)

7. 1 sign change one positive real zero.
Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros the polynomial may have. 1
starts Neg changes Pos.
1 sign change one positive real zero.
starts Pos Never changes
0 sign change no negative real zeros.
Descartes rule says one positive and no negative real zeros so there must be 4 complex zeros for a total of 5. We’ll learn more about complex zeros later.

8. The Rational Zeros Theorem
Now let’s factor a polynomial: 1
The Rational Zeros Theorem
We can get a list of the POSSIBLE rational zeros
Both positives and negatives could work for factors 
1, 2
Factors of the constant 1
Factors of the leading coefficient

9. 
1, 2
Possible zeros to try, in this case  1 or  2 1
Let’s try 1
YES! It is a zero since the remainder is 0
We found a positive real zero so Descartes Rule tells us there is another one

10. We could try 2, the other positive possible
We could try 2, the other positive possible. IMPORTANT: 1 may work again because it could have a multiplicity. 
1, 2 1
Let’s try 1 again to try and factor the depressed polynomial
YES! the remainder is 0
Now we can factor using regular methods or use the quadratic formula.

11. There ended up being two positive real zeros, 1 and 1 and two negative real zeros, -2, and -1.
In this factored form we can find intercepts and left and right hand behavior and graph the polynomial
Plot intercepts
Left & right hand behavior
Touches at 1 crosses at -1 and -2.
“Rough” graph

12.  1, 3, 9 1, 2 Example: factors of constant
factors of leading coefficient
So possible rational zeros are:

13. Descartes Rule helps us narrow down the choices.
starts Pos. Stays positive
Descartes Rule helps us narrow down the choices.
No sign changes in f(x) so no positive real zeros
starts Pos. changes Neg. changes Pos. Changes Neg. Changes Pos. 1 2 3 4
4 sign changes so 4 or 2 or 0 negative real zeros.

14. Let’s try -1
Yes! We found a zero. Let’s work with reduced polynomial then.
Yes! We found a zero. Let’s work with reduced polynomial then.
Let’s try -1 again
Yes! We found another one. We are done with trial and error since we can put variables back in and solve the remaining quadratic equation.
So remaining zeros found by setting these factors = 0 are -3/2 and -3. Notice these were in our list of choices.

15. So our polynomial factored is:

16. Upper and Lower Bound Rule:
Let f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x – c, using synthetic division.
If c > 0 and each number in the 1st row is either positive or zero, c is an upper bound for the real zeros of f.
If c < 0 and the numbers in the last row are alternatively positive and negative (zero entries count as positive or negative), c is a lower bound for the real zeros of f.

17. Find the zeros:
By Descartes Rule: There are 3 sign changes in f(x) so there are 3 or 1 positive real zeros.
There are no sign changes in f(-x) so there 0 negative real zeros.
Using the rational zeros theorem, the possible rational zeros are:
1, 2
1, 2, 3, 6

18.
1 is not a zero and f(1) = 3
Let’s try 1
The last row has all positive entries!
x = 1 is an upper bound for the real zeros.
We can restrict the search by trying numbers in the list less than 1!
Eventually we try 2/3 2/
2/3 is a zero and f(2/3) = 0
Thus,

19. Sneedlegrit: Find all rational zeros:
HW: pg. 248(2 – 14even, 18, 20, 26, 31, 46, 48)