1. Objective: To determine the end behavior of polynomial functions
Objective: To determine the end behavior of polynomial functions. To factor higher degree polynomials using the factoring by grouping method.
Polynomial Functions

2. Polynomial Functions
For each of the functions and their given graphs below, determine:
The degree of the function
Whether it is even or odd
The number of zeros
The number of x-intercepts
The number of turns
The leading coefficient
Whether the ends go in the same or in opposite directions
The end behavior of the function as x  ∞

3. Polynomial Functions f(x) = 2x3+5x2 – 18x – 15 Degree: Even or odd:
# of zeros:
# of x-intercepts:
# of turns:
Leading coefficient:
Ends same or opposite:
Behavior as x  ∞ : 3
odd 3 3 2 2
opposite
y  ∞

4. Polynomial Functions f(x) = x4 – 2x3 – 13x2 + 14x + 24 Degree:
Even or odd:
# of zeros:
# of x-intercepts:
# of turns:
Leading coefficient:
Ends same or opposite:
Behavior as x  ∞ : 4
even 4 4 3 1
same
y  ∞

5. Polynomial Functions f(x) =-2x5 – 5x4 + 24x3 + 41x2 – 34x – 24 Degree:
Even or odd:
# of zeros:
# of x-intercepts:
# of turns:
Leading coefficient:
Ends same or opposite:
Behavior as x  ∞ : 5
odd 5 5 4 -2
opposite
y  -∞

6. Polynomial Functions Degree: Even or odd: # of zeros:
f(x) =-0.5x x x4 – 43.5x3 – 176x x + 360
Degree:
Even or odd:
# of zeros:
# of x-intercepts:
# of turns:
Leading coefficient:
Ends same or opposite:
Behavior as x  ∞ : 6
even 6 6 5
-0.5
same
y  -∞

7. Polynomial Functions
From what we have seen in the previous examples, we can conclude the following:
The number of zeros is equal to the degree of the function
The maximum number of x-intercepts is equal to the degree of the function
The maximum number of turns is 1 less than the degree
The ends of the function go in the same direction for even functions and opposite directions for odd functions
The sign of the leading coefficient tells us if the right-end of the function will go up or down

8. Polynomial Functions
A function with ODD degree will always have an ODD number of real zeros.
A function with EVEN degree will always have an even number of real zeros, or will have no real zeros at all.
IMAGINARY/COMPLEX zeros will ALWAYS occur in conjugate pairs. (3 + 4i and 3 – 4i, for example)

9. Polynomial Functions
Use what you have just learned to determine the following for:
f(x) = x8 + x 7+ 6x6 – 5x4 + 3x - 2
Degree:
Even or odd:
# of zeros:
Max # of turns:
Max # of x-int:
Leading coefficient:
Ends same or opposite:
Behavior as x  ∞ :
Possible number of imaginary zeros:
Minimum number of real zeros: 8
even 8 7 8 1
same
y  ∞
0, 2, 4, 6 or 8

10. Polynomial Functions
Use what you have just learned to determine the following for:
f(x) = 3 – 2x4 + 7x2 – 3.5x + 2.5x3 – 0.25x7 + 5x5
Degree:
Even or odd:
# of zeros:
Max # of turns:
Max # of x-int:
Leading coefficient:
Ends same or opposite:
Behavior as x  ∞ :
Possible number of imaginary zeros:
Minimum number of real zeros: 7
odd 7 6 7
-0.25
opposite
y  -∞
0, 2, 4 or 6 1

11. Factoring by Grouping
Factoring by Grouping can be used to factor some 3rd degree and higher polynomials
Factoring by Grouping groups the polynomial into two or more parts and uses the GCF to complete the process.
In Factoring by Grouping, the GCF is often a binomial.

12. Factoring by Grouping Factor: x3 – x2 + x – 1
In this problem, we can group the first two and the last two terms
(x3 – x2) + (x – 1)
The first group has a GCF of x2 and the second group has a GCF of 1
x2(x – 1) + 1(x – 1)
Now, we have a GCF of (x – 1)
(x – 1)(x2 + 1)
Neither binomial can factor any more, so this is the final answer

13.  This is a difference of cubes!!
Factoring by Grouping
Factor: x4 + 2x3 – x2 – 2
Group the first two and the last two terms
(x4 + 2x3) + (-x2 – 2)
The first group has a GCF of x3 and the second group has a GCF of -1
x3(x + 2) – 1(x + 2)
Now each group has a GCF of (x + 2)
(x + 2)(x3 – 1)
(x + 2)(x – 1)(x2 + x + 1)
 This is a difference of cubes!!

14. Factoring by Grouping Factor: x3 + 7x2 – 4x – 28
For this problem, let’s group the 1st and 3rd terms and the 2nd and 4th terms
(x3 – 4x) + (7x2 – 28)
The first group has a GCF of x and the second group has a GCF of 7
x(x2 – 4) + 7(x2 – 4 )
Now each group has a GCF of (x2 – 4)
(x2 – 4)(x + 7)
(x + 2)(x – 2)(x + 7)
 (x2 – 4) is a difference of squares!!!

15. Factoring by Group Practice: a3 – 2a2 + 5a – 10 3x3 + 21x2 – 2x – 14

16. Factoring by Grouping 4x3 + 12x2 – 9x – 27 (4x3 + 12x2) + (-9x – 27)