Objective To determine the end behavior of polynomial

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

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 ∞

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

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

Polynomial Functions f(x) =-2 x 5 – 5 x 4 + 24 x 3 + 41 x 2 – 34 x – 24 Degree: 5 Even or odd: odd # of zeros: 5 # of x-intercepts: 5 # of turns: 4 Leading coefficient: -2 Ends same or opposite: opposite Behavior as x ∞ : y -∞

Polynomial Functions f(x) =-0. 5 x 6 + 1. 5 x 5 + 20. 5 x 4 – 43. 5 x 3 – 176 x 2 + 198 x + 360 Degree: 6 Even or odd: even # of zeros: 6 # of x-intercepts: 6 # of turns: 5 Leading coefficient: -0. 5 Ends same or opposite: same Behavior as x ∞ : y -∞

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 rightend of the function will go up or down

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 + 4 i and 3 – 4 i, for example)

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

Polynomial Functions Use what you have just learned to determine the following for: f(x) = 3 – 2 x 4 + 7 x 2 – 3. 5 x + 2. 5 x 3 – 0. 25 x 7 + 5 x 5 Degree: 7 Even or odd: odd # of zeros: 7 Max # of turns: 6 Max # of x-int: 7 Leading coefficient: -0. 25 Ends same or opposite: opposite Behavior as x ∞ : y -∞ Possible number of imaginary zeros: 0, 2, 4 or 6 Minimum number of real zeros: 1

Factoring by Grouping can be used to factor some 3 rd 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.

Factoring by Grouping Factor: x 3 – x 2 + x – 1 In this problem, we can group the first two and the last two terms (x 3 – x 2) + (x – 1) The first group has a GCF of x 2 and the second group has a GCF of 1 x 2(x – 1) + 1(x – 1) Now, we have a GCF of (x – 1)(x 2 + 1) Neither binomial can factor any more, so this is the final answer

Factoring by Grouping Factor: x 4 + 2 x 3 – x – 2 Group the first two and the last two terms (x 4 + 2 x 3) + (-x – 2) The first group has a GCF of x 3 and the second group has a GCF of -1 x 3(x + 2) – 1(x + 2) Now each group has a GCF of (x + 2)(x 3 – 1) This is a difference of cubes!! (x + 2)(x – 1)(x 2 + x + 1)

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

Factoring by Group Practice: a 3 – 2 a 2 + 5 a – 10 (a 3 – 2 a 2) + (5 a – 10) a 2(a – 2) + 5(a – 2) (a – 2)(a 2 + 5) 3 x 3 + 21 x 2 – 2 x – 14 (3 x 3 + 21 x 2) + (-2 x – 14) 3 x 2(x + 7)-2(x + 7) (x + 7)(3 x 2 – 2)

Factoring by Grouping 4 x 3 + 12 x 2 – 9 x – 27 (4 x 3 + 12 x 2) + (-9 x – 27) 4 x 2(x + 3) – 9(x + 3) (4 x 2 – 9) (x + 3)(2 x – 3)