1. Exponents, Polynomials, and Polynomial Functions
Chapter 6
Exponents, Polynomials, and Polynomial Functions
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9 1

2. Exponents and Scientific Notation
6.1
Exponents and Scientific Notation
Objectives:
Use the product & quotient rules for exponents
Evaluate expressions raised to the 0 power
Evaluate expressions raised to the negative nth power
Convert between scientific notation and standard notation

3. Product Rule for Exponents If m and n are positive integers and a is a real number, then am · an = am+n
Zero Exponent
If a does not equal 0, then a0 = 1.
Quotient Rule for Exponents
If a is a nonzero real number and n and m are integers, then
Negative Exponents
If a is a real number other than 0 and n > 0, then

4. Example 1 Use the product rule to simplify. a. 32 · 34 = 32+4 = 36
b. z3 · z2 · z5
= z3+2+5
= z10
c. (3y2)(–4y4)
= 3(–4)(y2 · y4)
= –12y6

5. Example 2 Evaluate the following. a. 50 = 1 b. −50 = − (50) = − (1)
= −1
Without parentheses, only 5 is raised to the 0 power.
c. (3x + 4)0
= 1 d.
–x0
= –(x0)
= –1

6. Example 3
Use the quotient rule to simplify. a. b.
= 3 a4 – 1 b7 - 2

7. Example 4 Simplify and write with positive exponents only. a. 3−2
b. 2x−4
Without parentheses, only x is raised to the –4 power. c. d.
= 7p8p2
= 7p10

8. Scientific Notation
In many fields of science we encounter very large or very small numbers. Scientific notation is a convenient shorthand for expressing these types of numbers.
A positive number is written in scientific notation if it is written as a product of a number a, where 1  a < 9, and an integer power r of 10.
a  10r
Scientific notation

9. Example 5 Write each number in scientific notation. 4700
Move the decimal 3 places to the left, so that the new number has a value between 1 and 10.
4700 a.
Since we moved the decimal 3 places, and the original number was > 10, our count is positive 3.
4700 = 4.7  103
Move the decimal 4 places to the right, so that the new number has a value between 1 and 10. b.
Since we moved the decimal 4 places, and the original number was < 1, our count is negative 4.
= 4.7  10−4

10. Example 6
Write each number in standard notation.
a  103
Since the exponent is a positive 3, we move the decimal 3 places to the right. =
Since the exponent is a negative 5, we move the decimal 5 places to the left.
6.45  10-5 b.
 10-5 =

11. 6.1 summary Objectives: Use the product & quotient rules for exponents
Evaluate expressions raised to the 0 power
Evaluate expressions raised to the negative nth power
Convert between scientific notation and standard notation

12. More Work with Exponents and Scientific Notation
6.2
More Work with Exponents and Scientific Notation
Objectives:
Use the power rules for exponents
Use exponents rules and definitions to simplify exponential expressions
Compute using scientific notation

13. The Power Rule and Power of a Product or Quotient Rule for Exponents If a and b are real numbers and m and n are integers, then
(am)n = amn
Power rule
(ab)n = an · bn
Power of a product
Power of a quotient

14. Example 1
Use the power rule to simplify the following expressions. Use positive exponents to write all results.
a. (x7)2
= x7•2
= x14
b. (23)3
= 23·3
= 29
= 512
c. (3−1)3
= 3−1•3
= 3−3

15. Example 2
Use the power rule to simplify the following expressions. Use positive exponents to write all results.
a. (5x2y)3
= 53 · (x2)3 · y3
= 125x6 y3

16. Summary of Rules for Exponents
If a and b are real numbers m and n are integers, then:
Product rule am · an = am+n
Zero exponent a0 = (a  0)
Negative exponent
Quotient rule
Power rule (am)n = amn
Power of a product (ab)m = am · bm
Power of a quotient

17. Example 3
Simplify each expression. Use positive exponents to write the answers.
a. (4xy−4)−2
= 4−2x−2(y−4)−2
= 4−2x−2y8

18. Example 4
Simplify the expression. Use positive exponents to write the answer.

19. Example 5
Simplify each expression. Assume that a and b are integers and that x and y are not 0. a.
= x-b 43 x3b
= 64 x-b+3b
= 64x2b
= y8a–(a-4)
= y8a-a+4
= y7a+4

20. Example 6
Perform the following operations. Write each result in scientific notation.
(7.3  102)(8.1  105) a.
= (7.3 · 8.1)  (102 · 105)
=  103
= 5.913x104 b.
= 3x10-6

21. Example 7 Use scientific notation to simplify
Write the result in scientific notation.

22. 6.2 summary Objectives: Use the power rules for exponents
Use exponents rules and definitions to simplify exponential expressions
Compute using scientific notation

23. Polynomials and Polynomial Functions
6.3
Polynomials and Polynomial Functions
Objectives:
Define a polynomial
Combine like terms
Add/subtract polynomials
Recognize the graph of a polynomial from the degree of the polynomial

24. Polynomial Vocabulary
Term – a number or the product of a number and one or more variables raised to powers
Coefficient – numerical factor of a term
Constant – term which is only a number
Polynomial is a sum of terms involving variables raised to a whole number exponent, with no variables appearing in any denominator.
Types of Polynomials
Monomial is a polynomial with one term.
Binomial is a polynomial with two terms.
Trinomial is a polynomial with three terms.

25. More Polynomial Vocabulary
The degree of a term is the sum of the exponents on the variables contained in the term.
The degree of a polynomial is the largest degree of all its terms
In the polynomial 7x5 + x2y2 – 4xy + 7
What are the terms:
The coefficients are:
Which is the constant term?
Degree of each term is:
Degree of the polynomial is
7x5, x2y2, –4xy and 7
7, 1, –4, and 7 7
5, 4, 2, and 0 5

26. Example 1 Find the degree of each term. a. 5x7 b. −43x6 c. 7.26
The exponent on x is 7, so the degree of the term is 7.
b. −43x6
The exponent on x is 6, so the degree of the term is 6.
c. 7.26
The degree of 7.26, which can be written as 7.26x0, is 0.

27. Example 2
Find the degree of each polynomial and indicate whether the polynomial is also a monomial, binomial, or trinomial.
Polynomial
Degree
Classification 2
Trinomial 4
Monomial 6
Binomial

28. Example 3 Find the degree of the polynomial Degree: 2 5 1 0
The largest degree of any term is 5, so the degree of this polynomial is 5.

29. Example 4 If P(x) = 4x2 − 3x + 1. Find the following. a. P(2)
Substitute 2 in for x and simplify.
P(2) = 4(2)2 − 3(2) + 1 = 11
b. P(−3)
Substitute −3 in for x and simplify.
P(−3) = 4(−3)2 − 3(−3) + 1 = 46

30. Like terms are terms that contain exactly the same variables raised to exactly the same powers.
Unlike Terms
−7x2, −x2
5x2, −x
8x2y4z, 5x2zy4
6x2y3, −4xy2
Example 5:
Combine like terms to simplify.
x2y + xy – y + 10x2y – 2y + xy
= 11x2y + 2xy – 3y

31. Example 6 Simplify each: (3x – 8) + (4x2 – 3x +3)
(4 – 3y) – (– y – 4)
= 4 – 3y + y + 4
= –2y + 8
(– a2 + 1) – (a2 – 3) + (5a2 – 6a + 7)
= –a2 + 1 – a a2 – 6a + 7
= 3a2 – 6a + 11

32. Types of Polynomials
Using the degree of a polynomial, we can determine what the general shape of the function will be, before we ever graph the function.
A polynomial function of degree 1 is a linear function.
A polynomial function of degree 2 is a quadratic function.
In general, for the quadratic equation of the form y = ax2 + bx + c, and the graph is a parabola.
a > 0 x
a < 0 x

33. Polynomial functions of degree 3 are cubic functions
Polynomial functions of degree 3 are cubic functions. Cubic functions have an x3 term.
x3 coefficient is negative
x3 coefficient is positive

34. Find the time when the object hits the ground
Add on HW help:
Projectile question
s(t) = -16t t
Find the time when the object hits the ground
Hits the ground when the height = 0
Solve for t when s(t) = 0
0 = -16t t
0 = -t(16t – 681)
t = or t = 681/16 = seconds
The object hits the ground in seconds

35. 6.3 summary Objectives: Define a polynomial Combine like terms
Add/subtract polynomials
Recognize the graph of a polynomial from the degree of the polynomial
Vocabulary:
Polynomial
Constant term
Monomial, Binomial, & Trinomial

36. Multiplying Polynomials
6.4
Multiplying Polynomials
Objectives:
Multiply two polynomials
Square binomials
Multiply the sum and difference of two terms
Multiply three or more polynomials
Evaluate polynomial functions

37. Example 1 Multiply. a. (3x2)(–2x) = (3)(–2)(x2)(x) = –6x3
b. (6y3z2)(–xy5z4)
= (6)(–1)x(y3y5)(z2z4)
= –6xy8z6
c. (4x2)(3x2 – 2x + 5)
= 12x4 – 8x3 + 20x2

38. Example 2 Multiply. (2x – 4)(7x + 5) = 2x(7x + 5) – 4(7x + 5)
–20

39. Example 3 (a + 2)(a3 – 3a2 + 7) Multiply.
= a(a3 – 3a2 + 7) + 2(a3 – 3a2 + 7)
= a4 – 3a3 + 7a + 2a3 – 6a2 + 14
= a4 – a3 – 6a2 + 7a + 14 a a4
+2a3
a3 –3a2 +7
-3a3
-6a2
+7a
+14

40. Example 4 Multiply (3x + 4)2 (5x – 2z)2 = (3x + 4)(3x + 4)
= (3x)(3x + 4) + 4(3x + 4)
= 9x x + 12x + 16
= 9x x + 16
(5x – 2z)2
= (5x – 2z)(5x – 2z)
= (5x)(5x – 2z) – 2z(5x – 2z)
= 25x2 – 10xz – 10xz + 4z2
= 25x2 – 20xz + 4z2

41. Example 5 Multiply (2x2 + x – 1)(x2 + 3x + 4)
= (2x2)(x2 + 3x + 4) + x(x2 + 3x + 4) – 1(x2 + 3x + 4)
= 2x4 + 6x3 + 8x2 + x3 + 3x2 + 4x – x2 – 3x – 4
= 2x4 + 7x3 + 10x2 + x – 4
Personally: I would have used a 3x3 box for this problem …

42. Special Products
In the process of using the FOIL method on products of certain types of binomials, we see specific patterns that lead to special products.
Square of a Binomial
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
Product of the Sum and Difference of Two Terms
(a + b)(a – b) = a2 – b2

43. Example 6 Multiply. a. (x – 6)2 = x2 – 2 • x • 6+ 62 = x2 – 12x + 36
b. (5x + 3z)2
= (5x)2 + 2(5x•3z) + (3z)2
= 25x2 + 30xz + 9z2
c. (x – 2)(x + 2)(x2 − 4)
(x – 2)(x + 2)(x2 − 4) = (x2 – 4)(x2 − 4)
= (x2 − 4)2
= x4 − 8x2 + 16

44. Example 7 If f(x) = 2x2 + 3x – 4, find f(a + 3).
f(a + 3) = 2(a + 3)2 + 3(a + 3) – 4
= 2(a2 + 6a + 9) + 3a + 9 – 4
= 2a2 + 12a a + 9 – 4
= 2a2 + 15a + 23

45. 6.4 summary Objectives: Multiply two polynomials Square binomials
Multiply the sum and difference of two terms
Multiply three or more polynomials
Evaluate polynomial functions

46. The Greatest Common Factor and Factoring by Grouping
6.5
The Greatest Common Factor and Factoring by Grouping
Objectives:
Identify the GCF
Factor out the GCF of a polynomial’s term
Factor polynomials by grouping

47. Example 1 Find the GCF of each list of terms. a. x3 and x7
b. 6x5 and 4x3
6x5 = 2 · 3 · x · x · x
4x3 = 2 · 2 · x · x · x
GCF is 2 · x · x · x = 2x3
x3 = x · x · x
x7 = x · x · x · x · x · x · x
GCF is x · x · x = x3

48. Example 2 Find the GCF of a3b2, a2b5 and a4b7.
a3b2 = a · a · a · b · b
a2b5 = a · a · b · b · b · b · b
a4b7 = a · a · a · a · b · b · b · b · b · b · b
GCF is a · a · b · b = a2b2

49. The first step in factoring a polynomial is to find the GCF of all its terms.
Then we write the polynomial as a product by factoring out the GCF from all the terms.

50. Example 3 Factor 6x3 – 9x2 + 12x = 3x · 2x2 – 3x · 3x + 3x · 4
Factor 14x3y + 7x2y – 7xy
= 7xy · 2x2 + 7xy · x – 7xy · 1
= 7xy(2x2 + x – 1)

51. Example 4 Factor. a. 6(x + 2) – y(x + 2)
The GCF is the binomial (x + 2)
= 6(x + 2) – y(x + 2)
= (x + 2)(6 – y)
b. xy(y + 1) – (y + 1)
The GCF is the binomial (y + 1)
= xy(y + 1) – 1(y + 1)
= (y + 1)(xy – 1)

52. Sometimes it is possible to factor a polynomial by grouping the terms of the polynomial and looking for common factors in each group. This method is called factoring by grouping.
Example 5
Factor xy + y + 2x + 2
= (xy + y) + (2x + 2)
= y(x + 1) + 2(x + 1)
= (x + 1)(y + 2)

53. Example 6 Factor. x3 + 4x + x2 + 4 b. 2x3 – x2 – 10x + 5

54. Example 7 Factor 2x – 9y + 18 – xy.
Neither pair has a common factor (other than 1).
So, rearrange the order of the factors.
2x + 18 – 9y – xy
= 2x + 18 – 9y – xy
= 2(x + 9) – y(9 + x)
= 2(x + 9) – y(x + 9)
= (x + 9)(2 – y)

55. 6.5 summary Do #9,10, 6, &1 now Objectives: Identify the GCF
Factor out the GCF of a polynomial’s term
Factor polynomials by grouping
Vocabulary:
Greatest common factor
Factoring by grouping
#37 & 38 Geometry
Formula for a Circle
(x – h)2 + (y – k)2 = r2
Center (h,k) and radius = r
Example for #38
X2 + Y2 = 9
(X+0)2 + (Y+0)2 = 9
then center (0,0) radius = 3

56. Factoring Trinomials 6.6 Objectives: Factor: x2 + bx + c
Factor: ax2 + bx + c
Factor by substitution

57. Example 1
Factor coefficient of x2 = 1 (Mrs. Hall’s call’s these quick factor)
x2 + 13x + 30
x2 – 2x – 35
= (x + 3)(x + 10)
= (x + 5)(x – 7)
+30
-35
1,30
2,15
3,10
5,6
-1,-30
-2,-15
-3,-10
-5,-6
1,-35
5,-7
-1,35
-5,7 +3
+10 +5 -7
+13 -2
x2 – 11x + 24
x2 – 6x + 10
= (x – 3)(x – 8)
prime polynomial, not factorable
+24
1,24
2,12
3,8
4,6
-1,-24
-2,-12
-3,-8
-4,-6
1,10
2,5
-1,-10
-2,-5
+10 -3 -8
-11 -6

58. Example 2
Factor
25x2 + 20x + 4
Mult (FL) = +100
with a sum of +20
1,100
2,50
4,25
5,20
10,10
-1,-100
-2,-50
-4,-25
-5,-20
-10,-10
= 25x2 + 10x + 10x + 4
= 5x(5x + 2) + 2(5x + 2)
= (5x + 2)(5x + 2) = (5x + 2)2
21x2 – 41x + 10
= 21x2 – 6x – 35x + 10
Mult (FL) = +210
with a sum of -41
1,210
2,105
3,70
5,42
6,35
7,30
10,21
-1,-210
-2,-105
-3,-70
-5,-42
-6,-35
-7,-30
-10,-21
= 3x(7x – 2) – 5(7x – 2)
= (3x − 5)(7x − 2)

59. Example 3 Factor 3x2 – 7x + 6 Prime 6x2y2 – 2xy2 – 60y2
Mult (FL) = +18
with a sum of -7
1,18
2,9
3,6
-1,-18
-2,-9
-3,-6
3x2 – 7x + 6
Prime
6x2y2 – 2xy2 – 60y2
Mult (FL) = -90
with a diff of -1
-1,90
-2,45
-3,30
-5,18
-6,15
-9,10
1,-90
2,-45
3,-30
5,-18
6,-15
9,-10
= 2y2(3x2 – x – 30)
= 2y2(3x2 + 9x – 10x – 30)
= 2y2 [3x(x + 3) – 10(x + 3)]
= 2y2(3x – 10)(x + 3)

60. Factoring a Trinomial of the Form ax2 + bx + c by Grouping
1. Find the two numbers whose product is a • c and whose sum is b.
2. Write the term bx as a sum by using the factors found in Step 1.
3. Factor by grouping.
You should always check your factoring results by multiplying the factored polynomial to verify that it is equal to the original polynomial.
Many times you can detect computational errors or errors in the signs of your numbers by checking your results.

61. Sometimes complicated polynomials can be rewritten into a form that is easier to factor by using substitution.
We replace a portion of the polynomial with a single variable, hopefully creating a format that is familiar to us for factoring purposes.

62. Example 4 Factor (4r + 1)2 + 8(4r + 1) + 16.
The quantity (4r + 1) is in two of the terms of this polynomial.
Substitute x for (4r + 1), and the result is the following simpler trinomial.
(4r + 1)2 + 8(4r + 1) + 16
This is a quick factor because coef of x2 is 1
x x
Factors of +16
with a sum of +8
= (x + 4)(x + 4)
= (x + 4)2
We then have to replace the original variable to get
= (4r )2
= (4r + 5)2

63. 6.6 summary Objectives: Do the following NOW: Factor: x2 + bx + c
Factor: ax2 + bx + c
Factor by substitution
Do the following NOW:
#1,3,9,8,2

64. Factoring by Special Products
6.7
Factoring by Special Products
Objectives:
Factor a perfect square trinomial
Factor a difference of two squares
Factor the sum or difference of two cubes

65. Any trinomial that factors into a single binomial squared is called a perfect square trinomial.
Perfect Square Trinomials
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
A binomial is a difference of two squares when it is the difference of the square of some quantity a and the square of some quantity b.
Difference of Two Squares
a2 – b2 = (a + b)(a – b)

66. Example 1 Factor x2 + 8x + 16 = (x + 4)2 16x2 – 8xy + y2 = (4x – y)2
Notice:
the first term is a square: x2 = (x)(x)
the last term is a square: 16 = (4)(4)
and 8x = 2(x)(4)
= (x + 4)2
16x2 – 8xy + y2
Notice:
the first term is a square 16x2 = (4x)(4x)
the last term is a square y2 = (y)(y)
and 8xy = 2(4x)(y)
= (4x – y)2

67. Example 2 Factor the following x2 – 16 16x2 – 9y2 = (x – 4)(x + 4)
c. 9x2 – 4
= (3x – 2)(3x + 2)
16x2 – 9y2
= (4x – 3y)(4x + 3y)
d. x8 – y6
= (x4 – y3)(x4 + y3)

68. Example 3 Factor x4 – 81 = (x2 + 9)(x2 − 9) = (x2 + 9)(x + 3)(x − 3)

69. Example 4 Factor x2 + 20x + 100 – x4. (x2 + 20x + 100) – x4
Factoring by grouping comes to mind since the sum of the first three terms of this polynomial is a perfect square polynomial.
(x2 + 20x + 100) – x4
(x + 10)2 – x4
(x + 10 – x2)(x x2) =

70. a3 + b3 = (a + b) ( ???) a2 - ab + b2 a a3 - a2b + ab2 + b + a2b - ab2
Although the sum of two squares usually cannot be factored, the sum of two cubes, as well as the difference of two cubes, can be factored as follows.
Sum and Difference of Two Cubes
a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3
= (a + b) ( ???)
= (a + b)(a2 – ab + b2) a2
- ab
+ b2 a a3
- a2b
+ ab2
+ b
+ a2b
- ab2
+ b3

71. Example 5 a3 + b3 = (a + b)(a2 – a • b + b2) Example 5 company’s way
Factor x3 + 1
= (x + 1)(???)
= (x + 1)(x2 – x + 1) x2
- x
+ 1 x x3
- x2
+ x
+ 1
+ x2
- x
+ 1
Example 5 company’s way
Factor x3 + 1
a3 + b3 = (a + b)(a2 – a • b + b2)
x3 + 1 = x = (x + 1)(x2 – x • )
= (x + 1)(x2 – x + 1)

72. Factor x3 − 27 = (x − 3)(x2 + x • 3 + 32) = (x − 3)(x2 + 3x + 9)
Example 6
a3 − b3 = (a − b)(a2 + ab + b2)
Factor
x3 − 27
substitute x for a and 3 for b
= (x − 3)(x2 + x • )
= (x − 3)(x2 + 3x + 9)
Example 7
Factor 64y3 – x3y3
a3 − b3 = (a − b)(a2 + ab + b2)
= y3(64 – x3)
substitute 4 for a and x for b
= y3(4 – x)[42 + (4)(x) + (x2)]
= y3(4 – x)(16+ 4x + x2)

73. 6.7 summary Do #1,3,6,9,10 NOW Perfect Square Trinomials
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
Do #1,3,6,9,10
NOW
Difference of Two Squares
a2 – b2 = (a + b)(a – b)
Sum and Difference of Two Cubes
a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab + b2)

74. Solving Equations by Factoring and Problem Solving
6.8
Solving Equations by Factoring and Problem Solving
Objectives:
Solve polynomial equations by factoring
Solve problems that can be modeled by polynomial equations
Find the x-intercepts of a polynomial function

75. Zero-Factor Property
If a and b are real numbers and a • b = 0, then a = 0 or b = 0. This property is true for three or more factors, also.
Solving Polynomial Equations by Factoring
Write the equation in standard form so that one side of the equation is 0.
Factor the polynomial completely.
Set each factor containing a variable equal to 0.
Solve the resulting equations.
Check each solution in the original equation.

76. Example 1 Solve: (x + 7)(x – 2) = 0 x + 7 = 0 or x – 2 = 0
Ans: {−7, 2}.
Solve: 2(x + 7)(x – 2) = 0
x + 7 = 0 or x – 2 = 0
Ans: {−7, 2}.
x = −7 or x = 2
Solve: 2x(x + 7)(x – 2) = 0
2x = 0 or x + 7 = 0 or x – 2 = 0
x = 0 or x = −7 or x = 2
Ans: {0,−7, 2}.

77. Example 2 Solve: 2x2 – 5x − 88 = 0 (2x + 11)(x − 8) = 0
-176
Solve: 2x2 – 5x − 88 = 0
+11
-16
= -8 -5
(2x + 11)(x − 8) = 0
2x + 11 = 0 or x − 8 = 0
Factors of -176
for a diff of -5
1,176
2,88
4,44
8,22
11,16
2x = −11 or
x = 8
Both solutions check so the solution set is

78. Example 3 Solve: 4x(8x + 9) = 5 32x2 + 36x = 5 32x2 + 36x − 5 = 0
-160 -4
+40
Solve: x(8x + 9) = 5
+36
32x2 + 36x = 5
Factors of -160
for a diff of 36
1,160
2,80
4,40
5,32
8,20
10,16
32x2 + 36x − 5 = 0
(8x − 1)(4x + 5) = 0
8x − 1 = 0 or 4x + 5 = 0
8x = 1
4x = − 5 or
Both solutions check so the solution set is

79. Example 4 Solve: 3(x2 + 4) + 5 = −6(x2 + 2x) + 13
I checked for perfect square since 9x2 and 4 are perfect squares
2•3x•2 = 12x so it is a perfect square
(3x + 2)(3x + 2) = 0
3x + 2 = 0 or 3x + 2 = 0
3x = −2
The solution is

80. Example 5 Solve: x3 = 16x. x3 − 16x = 0 x(x2 − 16) = 0
x = 0 or x + 4 = 0 or x − 4 = 0
x = 0 or x = −4 or x = 4
The solution are −4, 0, and 4
Check by substituting into the original equation.

81. Example 6
The height, h, in feet of a golf ball after t seconds is given by h = −16t2 + 96t. Find how long it takes the golf ball to return to the ground.
When t = 1 second, the height is
h = −16(1)2 + 96(1) = 80 feet
When t = 2 seconds, the height is
h = −16(2)2 + 96(2) = 128 feet
We want to know what value of t makes the height h equal to 0. We want to solve h = 0.
−16t2 + 96t = 0
−16t(t – 6) = 0
−16t = 0 or t – 6 = 0
The golf ball returns to the ground 6 seconds after it is hit.
t = 0 or t = 6

82. Example 7
The product of two consecutive positive integers is Find the two integers.
x = one of the unknown positive integers
x + 1 = the next consecutive positive integer.
x(x + 1) = 132
-132
x2 + x = 132
+12
-11 +1
x2 + x – 132 = 0
(x + 12)(x – 11) = 0
x + 12 = 0 or x – 11 = 0
The two positive integers are 11 and 12.
x = −12 or x = 11

83. Example 8
Find the length of the shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg.
x2 + (x + 10)2 = (2x – 10)2
x2 + x2 + 20x = 4x2 – 40x + 100
2x2 + 20x = 4x2 – 40x + 100
2x2 – 60x = 0
2x(x – 30) = 0
x = 0 or x = 30
The length of the shorter leg is 30 miles.

84. Solutions of a polynomial equation are the x-intercepts of the graph of the related function and the x-intercepts of the graph of a polynomial function are the solution of the related polynomial equation. These values are also called roots, or zeros, of the polynomial function.

85. Example 9 Match each function with its graph. A. B. C. Zeros are:
0, -3, 3
Zeros are:
-3, 3, 1
Zeros are:
-3, 1
Graph B
Graph C
Graph A A. B. C.

86. Ex 10 – add on to notes
And then finish like the first section of these notes

87. 6.8 summary Do NOW #1,5,10,15,16,17 Ex #5: Story Problems #21,22,23,24
Objectives:
Solve polynomial equations by factoring
Solve problems that can be modeled by polynomial equations
Find the x-intercepts of a polynomial function
Do NOW
#1,5,10,15,16,17
Ex #5:
Story Problems
#21,22,23,24

88. Even & Odd Functions and End Behavior
6.9
Even & Odd Functions and End Behavior
Objectives:
Graph even and odd power functions
Describe the end behavior of a polynomial function.

89. State if each is an even or an odd function
An even function is a function symmetric to the y-axis.
Switch (-x) for x and yields the same equation
An odd function is a function symmetric to the x-axis.
Switch (-x) in function notation and yields the opposite equation.
Example 1
State if each is an even or an odd function
F(x) = 2x3
G(x) = -3x2
H(x) = 2x3+ 2x
J(x) = 2x3 – 5
F(-x) = 2(-x)3
G(-x) = -3(-x)2
H(-x) = 2(-x)3+ 2(-x)
J(-x) = 2(-x)3 – 5
= -2x3
odd
even
neither
= -3x2
= -2x3 – 2x
= -2x3 – 5

90. Discuss end behavior with several problems on the whiteboard
Have the students stand up and use their arms to demonstrate end behavior
f(x) = 2x3
g(x) = -3x2
h(x) = -2x3+ 2x
j(x) = 2x4 – 5
NO calculator for this slide

91. Describe the end behavior for the graph of each function
Example 2
Describe the end behavior for the graph of each function
f(x) = 2x3 as x → ∞, f(x) →
x →-∞, f(x) →
g(x) = -3x2 as x → ∞, g(x) →
x →-∞, g(x) →
h(x) = -2x3+ 2x as x → ∞, h(x) →
x →-∞, h(x) →
j(x) = 2x4 – 5 as x → ∞, j(x) →
x →-∞, j(x) → ∞ -∞

92. 6.9 summary NO calculator for this homework
You may use a calculator for the Geometry questions
Objectives:
Graph even and odd power functions
Describe the end behavior of a polynomial function.
Geometry Help:
#27 equation of circle is (x – h)2 + (y – k)2 = r2
They give you the end points of the diameter
The midpoint will be the center of the circle (h,k)
The distance from the radius to the endpoint will be the radius
Endpoints (2,-3) and (-6,-13) so the center is (-2,-8)
Radius = distance from (2,-3) to (-2,-8)
Center (-2,-8) and radius of √41 Ans: (x + 2)2 (y + 8)2 = 41
Do NOW
#3,4,12,15,22
√41 5 4