1. Chapter 6 - Polynomial Functions
Algebra 2

2. Table of Contents 6.1 - Polynomials 6.2 - Multiplying Polynomials
6.3 - Dividing Polynomials
6.4 - Factoring Polynomials
6.5 - Finding Real Roots of Polynomial Equation

3. 6.1 - Polynomials
Algebra 2

4. 6.1
Algebra 2 (Bell work)
A monomial is a number or a product of numbers and variables with whole number exponents. Ex: 2x
A polynomial is multiple monomials or terms put together in an expression
ex: 3x2 + 2x + 1
**Polynomials have no variables in the denominators or exponents, no roots or absolute values of variables, and all variables have whole number exponents.** 1 2 a7
Polynomials:
3x4
2z12 + 9z3
0.15x101
3t2 – t3 8
5y2 1 2
Not polynomials: 3x
|2b3 – 6b|
m0.75 – m
The degree of a monomial is the sum of the exponents of the variables.

5. A. z6 B. 5.6 C. 8xy3 D. a2bc3 Identify the degree of each monomial. z6
6.1
Example 1
Identifying the Degree of a Monomial
Identify the degree of each monomial.
A. z6
B. 5.6 z6
Identify the exponent.
5.6 = 5.6x0
Identify the exponent.
The degree is 6.
The degree is 0.
C. 8xy3
D. a2bc3
8x1y3
Add the exponents.
a2b1c3
Add the exponents.
The degree is 4.
The degree is 6.

6. 6.1
An degree of a polynomial is given by the term with the greatest degree or power
Standard Form: When the expression is written by using highest powers in descending order
When in standard form with the highest degree first, the leading coefficient is the number in front of that highest power term

7. A polynomial can be classified by its number of terms
6.1
A polynomial can be classified by its number of terms
A polynomial with two terms is called a binomial
A polynomial with three terms is called a trinomial.
A polynomial can also be classified by its degree.

8. Math Joke Knock, Knock Who’s there? Polly Polly who?
6.1
Math Joke
Knock, Knock
Who’s there?
Polly
Polly who?
Polynomial, Why the third degree?

9. 6.1
Example 2
Classifying Polynomials
Rewrite each polynomial in standard form.
Then identify the leading coefficient, degree, and number of terms.
Name the polynomial.
A. 3 – 5x2 + 4x
B. 3x2 – 4 + 8x4
Write terms in descending order by degree.
Write terms in descending order by degree.
–5x2 + 4x + 3
8x4 + 3x2 – 4
Leading coefficient: –5
Leading coefficient: 8
Degree: 2
Degree: 4
Terms: 3
Terms: 3
Name: quadratic trinomial
Name: 4th degree trinomial

10. 6.1
Optional
Rewrite each polynomial in standard form.
Then identify the leading coefficient, degree, and number of terms.
Name the polynomial.
a. 4x – 2x2 + 2
b. –18x2 + x3 – 5 + 2x
Write terms in descending order by degree.
Write terms in descending order by degree.
–2x2 + 4x + 2
1x3– 18x2 + 2x – 5
Leading coefficient: –2
Leading coefficient: 1
Degree: 2
Degree: 3
Terms: 3
Terms: 4
Name: quadratic trinomial
Name: cubic polynomial with 4 terms

11. A. (2x3 + 9 – x) + (5x2 + 4 + 7x + x3) B. (3 – 2x2) – (x2 + 6 – x) 6.1
Example 3
Adding and Subtracting Polynomials
Add or subtract. Write your answer in standard form.
A. (2x3 + 9 – x) + (5x x + x3)
B. (3 – 2x2) – (x2 + 6 – x)
Add the opposite horizontally.
3x3 + 5x2 + 6x + 13
(3 – 2x2) – (x2 + 6 – x)
(–2x2 + 3) + (–x2 + x – 6)
(–2x2 – x2) + (x) + (3 – 6)
–3x2 + x – 3

12. 6.1
Add or subtract. Write your answer in standard form.
(–36x2 + 6x – 11) + (6x2 + 16x3 – 5)
(5x x2) – (15x2 + 3x – 2)
(–36x2 + 6x – 11) + (6x2 + 16x3 – 5)
(5x x2) – (15x2 + 3x – 2)
–36x2 + 6x – 11
(5x3 + 6x2 + 12) + (–15x2 – 3x + 2)
+16x3 + 6x – 5
(5x3) + (6x2 – 15x2) + (–3x) + (12 + 2)
16x3 – 30x2 + 6x – 16
5x3 – 9x2 – 3x + 14

13. B. A. f(x) = 2x3 – 3x Day 2 6.1 Example 5
Graphing Higher-Degree Polynomials on a Calculator
Day 2
Graph each polynomial function on a calculator. Describe the graph and identify the number of real zeros.
A. f(x) = 2x3 – 3x B.
From left to right, the graph alternately decreases and increases, changing direction 3 times and crossing the x-axis 4 times; 4 real zeros.
From left to right, the graph increases, then decreases, and increases again. It crosses the x-axis 3 times, so there appear to be 3 real zeros.

14. a. f(x) = 6x3 + x2 – 5x + 1 b. f(x) = 3x2 – 2x + 2 6.1 Optional
Graph each polynomial function on a calculator. Describe the graph and identify the number of real zeros.
a. f(x) = 6x3 + x2 – 5x + 1
b. f(x) = 3x2 – 2x + 2
From left to right, the graph increases, decreases slightly, and then increases again. It crosses the x-axis 3 times, so there appear to be 3 real zeros.
From right to left, the graph decreases and then increases. It does not cross the x-axis, so there are no real zeros.

15. The cost of manufacturing a certain product can be approximated
6.1
Example 4
Application
The cost of manufacturing a certain product can be approximated
by f(x) = 3x3 – 18x + 45, where x is the number of units of the product in hundreds
Evaluate f(0) and f(200) and describe what the values represent.
You can simply do algebra or can use your calculator
f(0) = 3(0)3 – 18(0) + 45 = 45
f(200) = 3(200)3 – 18(200) + 45 = 23,996,445
f(0) represents the initial cost before manufacturing any products.
f(200) represents the cost of manufacturing 20,000 units of the products.

16. HW pg. 410
6.1-
Day 1: 3-9 (Odd), 10-13, 27-30, 64-72
Day 2: (Calc, Sketch Graph), 45, 47-49
Ch: 41-44, 50-53, 59-63

17. 6.2 - Multiplying Polynomials
Algebra 2

18. To multiply any two polynomials
6.2
Algebra 2 (Bell work)
Just Read
To multiply any two polynomials
Use the Distributive Property and multiply each term in the second polynomial by each term in the first.

19. A. 4y2(y2 + 3) B. fg(f4 + 2f3g – 3f2g2 + fg3) Find each product.
6.2
Example 1
Multiplying a Monomial and a Polynomial
Find each product.
A. 4y2(y2 + 3)
B. fg(f4 + 2f3g – 3f2g2 + fg3)
fg(f4 + 2f3g – 3f2g2 + fg3)
4y2(y2 + 3)
fg  f4 + fg  2f3g – fg  3f2g2 + fg  fg3
4y2  y2 + 4y2  3
f5g + 2f4g2 – 3f3g3 + f2g4
4y4 + 12y2

20. a. 3cd2(4c2d – 6cd + 14cd2) b. x2y(6y3 + y2 – 28y + 30) 6.2
Find each product.
Optional
a. 3cd2(4c2d – 6cd + 14cd2)
b. x2y(6y3 + y2 – 28y + 30)
3cd2(4c2d – 6cd + 14cd2)
x2y(6y3 + y2 – 28y + 30)
3cd2  4c2d – 3cd2  6cd + 3cd2  14cd2
x2y  6y3 + x2y  y2 – x2y  28y + x2y  30
12c3d3 – 18c2d3 + 42c2d4
6x2y4 + x2y3 – 28x2y2 + 30x2y

21. Math Joke Q: When is a solution not an answer?
6.2
Math Joke
Q: When is a solution not an answer?
A:When you make it in a chemistry lab

22. 1. (a – 3)(2 – 5a + a2) 2. (3b – 2c)(3b2 – bc – 2c2) 6.2 Example 2
Multiplying Polynomials
Find the product.
1. (a – 3)(2 – 5a + a2)
2. (3b – 2c)(3b2 – bc – 2c2)
(a – 3)(a2 – 5a + 2)
a(a2) + a(–5a) + a(2) – 3(a2) – 3(–5a) –3(2)
a3 – 5a2 + 2a – 3a2 + 15a – 6
a3 – 8a2 + 17a – 6
(3b – 2c)(3b2 – 2c2 – bc)
3b(3b2) + 3b(–2c2) + 3b(–bc) – 2c(3b2) – 2c(–2c2) – 2c(–bc)
9b3 – 6bc2 – 3b2c – 6b2c + 4c3 + 2bc2
9b3 – 9b2c – 4bc2 + 4c3

23. 6.2
Notice the coefficients of the variables in the final product of (a + b)3. these coefficients are the numbers from the third row of Pascal's triangle.
Day 2
Pg. 416

24. [1(k)3(–5)0] + [3(k)2(–5)1] + [3(k)1(–5)2] + [1(k)0(–5)3]
6.2
Just Read
Pg. 416
A. (k – 5)3
(Don’t copy on next slide)
[1(k)3(–5)0] + [3(k)2(–5)1] + [3(k)1(–5)2] + [1(k)0(–5)3]
k3 – 15k2 + 75k – 125

25. A. (k – 5)3 [1(k)3(–5)0] + [3(k)2(–5)1] + [3(k)1(–5)2] + [1(k)0(–5)3]
6.2
Example 4
Expanding a Power of a Binomial
Method #1
Expand each expression
A. (k – 5)3
[1(k)3(–5)0] + [3(k)2(–5)1] + [3(k)1(–5)2] + [1(k)0(–5)3]
k3 – 15k2 + 75k – 125

26. 6.2
B. (6m – 8)3
Method #1
[1(6m)3(–8)0] + [3(6m)2(–8)1] + [3(6m)1(–8)2] + [1(6m)0(–8)3]
216m3 – 864m m – 512

27. c. (3x + 1)4 Expand each expression.
6.2
Example 5
Using Pascal’s Triangle to Expand Binomial Expressions
Method #2
Expand each expression.
c. (3x + 1)4
(3x)0 = 1 1 4 6 4 1
(3x)4
(3x)3
(3x)2
(3x)
(1)1
(1)2
(1)3
(1)4
[1(3x)4(1)0] + [4(3x)3(1)1] + [6(3x)2(1)2] + [4(3x)1(1)3] + [1(3x)0(1)4]
81x x3 + 54x2 + 12x + 1

28. 6.2
b. (x – 4)5
Method #2 1 5 10 10 5 1 X5 x4 x3 x2 x1
(-4)1
(-4)2
(-4)3
(-4)4
(-4)5
[1(x)5(–4)0] + [5(x)4(–4)1] + [10(x)3(–4)2] + [10(x)2(–4)3] + [5(x)1(–4)4] + [1(x)0(–4)5]
x5 – 20x x3 – 640x x – 1024

29. HW pg.418 6.2- Day 1: 1-7 (Odd), 21-25 (Odd), 70-74 (Even)
Day 2: 13 (Without Pascal’s), 15, 17, 31
Ch: 9, 26, 39, 52, 53, 55

30. 6.3 - Dividing Polynomials
Algebra 2

31. Just Read Algebra 2 (Bell work) 6.3
Polynomial long division is a method for dividing a polynomial by another polynomial of a lower degree.
It is very similar to dividing numbers.
Just Read

32. (–y2 + 2y3 + 25) ÷ (y – 3) (15x2 + 8x – 12) ÷ (3x + 1) 6.3 Example 1
Using Long Division to Divide Polynomials
Divide using long division.
(–y2 + 2y3 + 25) ÷ (y – 3)
(15x2 + 8x – 12) ÷ (3x + 1) 5x
+ 1
2y2
+ 5y
+ 15
3x x2 + 8x – 12
y – y3 – y2 + 0y + 25
–(15x2 + 5x)
–(2y3 – 6y2)
3x – 12
5y2 + 0y
–(3x + 1)
–(5y2 – 15y)
15y + 25
–13
–(15y – 45) 70
= 5x + 1 – 13
3x + 1
= 2y2 + 5y 70
y – 3

33. (x2 + 5x – 28) ÷ (x – 3) 6.3 Divide using long division Optional x + 8–4
= x + 8 – 4
x – 3

34. 6.3
Just Read

35. Math Joke Teacher: Why didn’t you do your homework?
6.3
Math Joke
Teacher: Why didn’t you do your homework?
Student: The long division took too long, and the synthetic division just wasn’t real

36. 1 3 (3x2 + 9x – 2) ÷ (x – ) (3x4 – x3 + 5x – 1) ÷ (x + 2) 6.3
Example 2
Using Synthetic Division to Divide by a Linear Binomial
Divide using synthetic division 1 3
(3x2 + 9x – 2) ÷ (x – )
(3x4 – x3 + 5x – 1) ÷ (x + 2) 1 3 –2 –2
– –1 –6 14
–28 46 1 1 3 3 –7 14
–23 45 1 3 3 10
3x3 – 7x2 + 14x – 23 + 45
x + 2 3x 1 3
x –

37. (6x2 – 5x – 6) ÷ (x + 3) 6.3 Divide using synthetic division Optional–3
–5 –6
–18 69 63 6
–23
Step 4 Write the quotient.
6x – 23 + 63
x + 3

38. Copy Entire Know it Note Below
6.3
Day 2
Copy Entire Know it Note Below

39. 6.3
Example 3
Using Synthetic Substitution
Use synthetic substitution to evaluate the polynomial for the given value. 1 3
a. P(x) = 2x3 + 5x2 – x + 7 for x = 2
b. P(x) = 6x4 – 25x3 – 3x + 5 for x = – 2 – – 1 3
6 – – 4 18 34 2 9 –3 –2 2 9 17 41 –6 6
–27 9 7
P(2) = 41
Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7.
P( ) = 7 1 3
P(2) = 2(2)3 + 5(2)2 – (2) + 7
P(2) = 41 

40. 6.3
Example 4
Application
Optional
Write an expression that represents the area of the top face of a rectangular prism when the height is x + 2 and the volume of the prism is x3 – x2 – 6x.
The volume V is related to the area A and the height h by the equation V = A  h. Rearranging for A gives A = V h
x3 – x2 – 6x
x + 2
A(x) = –2
1 –1 –6 0 –2 6 1 –3
The area of the face of the rectangular prism can be represented by A(x)= x2 – 3x.

41. 6.3
Write an expression for the length of a rectangle with width y – 9 and area y2 – 14y + 45.
The area A is related to the width w and the length l by the equation A = l  w.
y2 – 14y + 45
y – 9
l(x) = 9 – 9
–45 1 –5
The length of the rectangle can be represented by l(x)= y – 5.

42. 6.3
HW pg.426
6.3-
Day 1: 2-7, (Even), 39, 40
Day 2: 9-12, 27-30, 65-72
Ch: 32, 57-63

43. 6.4 - Factoring Polynomials
Algebra 2

44. 6.4
Algebra 2(bell work)

45. A. (x + 1); (x2 – 3x + 1) B. (x + 2); (3x4 + 6x3 – 5x – 10) 6.4
Example 1
Determining Whether a Linear Binomial is a Factor
Determine whether the given binomial is a factor of the polynomial P(x).
A. (x + 1); (x2 – 3x + 1)
B. (x + 2); (3x4 + 6x3 – 5x – 10) –2
– –10 –1 – –6 10 –1 4 3 –5 1 –4 5
P(–1) = 5
P(–2) = 0
so (x + 2) is a factor of
P(x) = 3x4 + 6x3 – 5x – 10.
P(–1) ≠ 0
so (x + 1) is not a factor of P(x) = x2 – 3x + 1.

46. a. (x + 2); (4x2 – 2x + 5) b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30)
6.4
Optional
a. (x + 2); (4x2 – 2x + 5)
b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30) –2 – 2
– –10 –8 20 2 4 10 4
–10 25 1 2 5
P(–2) = 25
P(2) = 0
so (3x – 6) is a factor of
P(x) = 3x4 – 6x3 + 6x2 + 3x – 30.
P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5.

47. Factor: x3 – x2 – 25x + 25. Factor: x3 – 2x2 – 9x + 18.
6.4
Example 2
Factoring by Grouping
Factor: x3 – x2 – 25x + 25.
Factor: x3 – 2x2 – 9x + 18.
(x3 – x2) + (–25x + 25)
(x3 – 2x2) + (–9x + 18)
x2(x – 1) – 25(x – 1)
x2(x – 2) – 9(x – 2)
(x – 1)(x2 – 25)
(x – 2)(x2 – 9)
(x – 1)(x – 5)(x + 5)
(x – 2)(x – 3)(x + 3)

48. 2x3 + x2 + 8x + 4. 6.4 (2x3 + x2) + (8x + 4) x2(2x + 1) + 4(2x + 1)

49. 6.4
Example 3
Factoring the Sum of Difference of Two Cubes
Day 2
Factor the expression.
4x x
125d3 – 8
4x(x3 + 27)
(5d)3 – 23
4x(x3 + 33)
(5d – 2)[(5d)2 + 5d  ]
4x(x + 3)(x2 – x  )
(5d – 2)(25d2 + 10d + 4)
4x(x + 3)(x2 – 3x + 9)

50. 8 + z6 2x5 – 16x2 6.4 Factor the expression. (2)3 + (z2)3 2x2(x3 – 8)
(2 + z2)[(2)2 – 2  z + (z2)2]
2x2(x – 2)(x2 + x  )
(2 + z2)(4 – 2z + z4)
2x2(x – 2)(x2 + 2x + 4)

51. 6.4
Math Joke
Student: The artists Picasso must have been really good at Algebra
Parent: Why do you say that?
Student: He was a famous cubist, so he probably had to do a lot of factoring

52. V(x) = x3 + 6x2 + 3x – 10. Example 4 Application 6.4
The volume of a plastic storage box is modeled by the function
V(x) = x3 + 6x2 + 3x – 10.
Identify the values of x for which V(x) = 0
Then use the graph to factor V(x).
V(x) has three real zeros at
x = –5, x = –2, and x = 1.
If the model is accurate, the box will have no volume if x = –5, x = –2, or x = 1.
One corresponding factor is (x – 1). 1
–10 1 7 10
V(x)= (x – 1)(x2 + 7x + 10) 1 7 10
V(x)= (x – 1)(x + 2)(x + 5)

53. 6.4
The volume of a rectangular prism is modeled by the function
V(x) = x3 – 8x2 + 19x – 12, which is graphed below.
Identify the values of x for which V(x) = 0, then use the graph to factor V(x).
V(x) has three real zeros at x = 1, x = 3, and x = 4.
If the model is accurate, the box will have
no volume if x = 1, x = 3, or x = 4.
One corresponding factor is (x – 1). 1
1 – –12 1 –7 12 1 –7 12
V(x)= (x – 1)(x2 – 7x + 12)
V(x)= (x – 1)(x – 3)(x – 4)

54. HW pg. 433 6.4- Day 1: 1-9 (Odd), 21-25 (Odd), 61, 63
Day 2: 10-16, 32 (Use Calc to find zeros), (Odd), 49
Ch: 16, 39, 40, 46-49

55. 6.5 - Finding Real Roots of Polynomial Equations
Algebra 2

56. 6.5
Example 1
Using Factoring to Solve Polynomial Equations
Solve the polynomial equation by factoring.
4x6 + 4x5 – 24x4 = 0
x = 26x2
4x4(x2 + x – 6) = 0
x4 – 26 x = 0
(x2 – 25)(x2 – 1) = 0
4x4(x + 3)(x – 2) = 0
(x – 5)(x + 5)(x – 1)(x + 1)
4x4 = 0 or (x + 3) = 0 or (x – 2) = 0
x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0
x = 0, x = –3, x = 2
x = 5, x = –5, x = 1 or x = –1
The roots are 0, –3, and 2.
The roots are 5, –5, 1, and –1.

57. 6.5
Solve the polynomial equation by factoring.
x3 – 2x2 – 25x = –50
#6 2x3 – 12x2 = 32x -192
X3 – 2x2 – 25x +50 = 0
X2( x – 2) – 25 ( x -2) = 0
(x2 -25)( x -2)
2x3 -12x2 -32x – 192 = 0
2(x3 – 6x2 -16x + 96) = 0
2[ x2(x – 6) – 16(x-6) ] =0
2[ (x-6)(x2 -16) ]= 0
2(x-6)(x-4)(x+4)=0
X=6,4,-4
(x + 5)(x – 2)(x – 5) = 0
x + 5 = 0, x – 2 = 0, or x – 5 = 0
x = –5, x = 2, or x = 5
The roots are –5, 2, and 5.

58. 6.5
Sometimes a polynomial equation has a factor that appears more than once
This creates a multiple root.
In 3x5 + 18x4 + 27x3 = 0 has two multiple roots, 0 and –3.
For example, the root 0 is a factor three times because 3x3 = 0.
The multiplicity of root r is the number of times that x – r is a factor of P(x). When a real root has even multiplicity, the graph of y = P(x) touches the x-axis but does not cross it.
When a real root has odd multiplicity greater than 1, the graph “bends” as it crosses the x-axis.

59. 6.5

60. 6.5
Example 3
Identifying All of the Real Roots of a Polynomial Equation
Identify the roots of each equation. State the multiplicity of each root.
x3 + 6x2 + 12x + 8 = 0
x3 + 6x2 + 12x + 8 = (x + 2)(x + 2)(x + 2)

61. 6.5
Identify the roots of each equation. State the multiplicity of each root.
x4 + 8x3 + 18x2 – 27 = 0
x4 + 8x3 + 18x2 – 27 = (x – 1)(x + 3)(x + 3)(x + 3)

62. Polynomial equations may also have irrational roots.
6.5
Polynomial equations may also have irrational roots.
Day 2

63. Math Joke Q: Why is the Rational Root Theorem so polite?
A: It minds its p’s and q’s

64. 6.5
#29 from book with calculator
x3 -7x2 + 14x – 6 = 0

65. 6.5
Identify the roots of each equation. State the multiplicity of each root.
x3 -7x2 + 14x – 6 = 0

66. Identify all the real roots of 2x3 - 9x + 2 = 0, using the rational zero test
6.5

67. # x3 + 6x2 – 5x – 30 = 0
6.5

68. HW pg. 442 6.5- Day 1:2-7, 15-19 (Odd), 49-53 (Odd)
Ch: 10, 23, 28, 35, 36-39
If notes take 3 days, Quiz will be on Tuesday