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Page 1: Unit 2: Polynomials Student Editioncremymath.weebly.com/.../3/22234558/unit_2_math_3_student_edition.pdfStudent Edition. Unit 2.1 Interpret the Structure of Expressions Student Learning

Unit 2: Polynomials

Student Edition

Page 2: Unit 2: Polynomials Student Editioncremymath.weebly.com/.../3/22234558/unit_2_math_3_student_edition.pdfStudent Edition. Unit 2.1 Interpret the Structure of Expressions Student Learning

Unit 2.1 Interpret the Structure of Expressions

Student Learning Targets (SWBAT):

Interpret parts of expressions such as terms, factors, coefficients

Use the structure of an expression to identify ways to rewrite it Add polynomials

Subtract polynomials

Multiply polynomials

Notes:

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Assignment 2.1:

Determine which are polynomial functions. State the degree, leading coefficient. For those

that are not functions explain why.

1. 𝑓(π‘₯) = 3π‘₯βˆ’2 + 17 2. 𝑓(π‘₯) = 2π‘₯5 βˆ’1

2π‘₯ + 9

3. 𝑓(π‘₯) = βˆ’9 + 2π‘₯ 4. β„Ž(π‘₯) = 13

5. π‘˜(π‘₯) = √27π‘₯ + 8π‘₯33

Add or subtract the following polynomials

6. (π‘₯4 + 8 βˆ’ 5π‘₯3) + (2π‘₯3 βˆ’ 2 βˆ’ 3π‘₯4) 7. (4π‘Ž2 βˆ’ 2 βˆ’ 8π‘Ž3) + (5π‘Ž2 βˆ’ 4π‘Ž + 8π‘Ž3)

8. (6π‘₯2 + 7π‘₯ + 3π‘₯3) βˆ’ (5π‘₯2 + 3π‘₯ + 7π‘₯3) 9. (8π‘₯3 βˆ’ 4π‘₯2 + 5π‘₯4) βˆ’ (5π‘₯3 βˆ’ 5π‘₯4 + 6π‘₯2)

10. (2 βˆ’ 3𝑖) + (6 + 5𝑖) 11. (√5 βˆ’ 3𝑖) + (βˆ’2 + βˆšβˆ’9)

12. (2 βˆ’ 3𝑖) + (3 βˆ’ 4𝑖) 13. (2 βˆ’ 𝑖) + (3 βˆ’ βˆšβˆ’3)

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Find each product

10. (5π‘š βˆ’ 6)(2π‘š + 7) 11. (7π‘Ÿ + 8)(3π‘Ÿ βˆ’ 5)

12. (2π‘₯ βˆ’ 4)2 13. (βˆšβˆ’4 + 𝑖)(6 βˆ’ 5𝑖)

14. (3𝑛3 + 2𝑛)(4𝑛2 βˆ’ 1) 15. (6π‘₯3 + 2π‘₯)(π‘₯ + 5)

16. (2π‘₯ + 3)3 17. (2 + 3𝑖)(2 βˆ’ 𝑖)

18. (2 βˆ’ 𝑖)(1 + 3𝑖) 19. (7𝑖 βˆ’ 3)(2 + 6𝑖)

Find the area and perimeter of the rectangles.

21. 22. (3π‘₯2 βˆ’ 2π‘₯ + 1)

(π‘₯ βˆ’ 3)

(3π‘₯2 + 4π‘₯ βˆ’ 1)

(2π‘₯ βˆ’ 4)

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2.2: Factoring Polynomials

Objectives:

- Factor expressions with a common factor

- Factor expressions when a = 1

- Factor expressions when a > 1

Notes:

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Assignment 2.2

Factor each expression completely.

1. 3π‘₯2 + 10π‘₯ + 3 2. 3π‘₯2 βˆ’ 13π‘₯ + 12

3. 9π‘₯3 + 3π‘₯2 βˆ’ 12π‘₯ 4. 2π‘₯3 βˆ’ π‘₯2 βˆ’ 3π‘₯

5. 14𝑒2 βˆ’ 33𝑒 βˆ’ 5

6. 3π‘₯4 + 7π‘₯3 + 2π‘₯2

7. π‘₯2 + 9π‘₯ + 14

8. 𝑦2 βˆ’ 11𝑦 + 30

9. 6𝑑2 + 5𝑑 + 1

10. 12π‘₯2 + 11π‘₯ βˆ’ 15

Solve each equation using factoring.

11. π‘₯2 + 9π‘₯ βˆ’ 22 = 0 12. 3π‘₯2 + 8π‘₯ + 4 = 0

13. π‘₯4 βˆ’ 29π‘₯2 + 100 = 0 14. 4π‘₯4 βˆ’ 14π‘₯2 + 12 = 0

15. π‘₯6 βˆ’ 9π‘₯4 βˆ’ π‘₯2 + 9 = 0

16. 4π‘₯4 βˆ’ 4π‘₯2 βˆ’ π‘₯2 + 1 = 0

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Unit 2.3 – Factoring Special Polynomials

Objectives:

- Factor expressions using grouping

- Factor over complex numbers

- Factor perfect cubes

- Factor perfect squares

Notes:

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2.3 Assignment

Factor by grouping.

1. 4π‘₯3 βˆ’ 20π‘₯2 + 4π‘₯ βˆ’ 20 2. π‘₯3 + 3π‘₯2 + 5π‘₯ + 15

3. π‘₯3 βˆ’ 4π‘₯2 + 5π‘₯ βˆ’ 20

4. π‘₯6 βˆ’ 3π‘₯4 + π‘₯2 βˆ’ 3

Factor the sum or difference of two cubes.

5. 𝑦3 βˆ’ 8 6. 𝑧3 + 64

7. 8π‘₯3𝑦3 βˆ’ 64π‘₯6 8. 64𝑧3 + 27

Factor the sum or difference of two squares.

9. 𝑧2 βˆ’ 49 10. 64 βˆ’ 25𝑦2

11. π‘₯2 + 9 12. 16𝑧2 + 25

13. 9𝑦2 βˆ’ 16 14. 36π‘₯2 + 64

Find the zeros of the function.

15. π‘₯3 + π‘₯2 + 4π‘₯ + 4 16. 9π‘₯2 βˆ’ 3π‘₯ βˆ’ 2

17. 5π‘₯3 βˆ’ 5π‘₯2 βˆ’ 10π‘₯ 18. π‘₯3 βˆ’ 25π‘₯

19. π‘₯6 βˆ’ 9π‘₯4 βˆ’ π‘₯2 + 9 20. 4π‘₯4 βˆ’ 4π‘₯2 βˆ’ π‘₯2 + 1

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2.4 Assignment! Mixed factoring…

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Unit 2.5 Graphing Polynomial Functions

Objectives:

- Graph polynomial function using the roots of the function

- Analyze properties of polynomial functions

Notes:

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Assignment 2.5

Identify the zeros for each of the following functions. State the multiplicity of each zero.

1. π’š = (𝒙 βˆ’ πŸ’)(𝒙 + πŸ“)πŸ–(𝒙 βˆ’ πŸ”)πŸ‘ 2. π’š = πŸπ’™πŸ•(𝒙 + πŸ–)𝟐(𝒙 βˆ’ πŸ—)πŸ‘

3. π’š = βˆ’πŸ’π’™πŸ“(𝒙 βˆ’ 𝟏)(πŸπ’™ + πŸ“)πŸ’ 4. π’š = (πŸ•π’™ + 𝟐𝟏)πŸ—(𝒙 βˆ’ πŸ’)πŸ‘(πŸπ’™ + 𝟏𝟎)

Graph each polynomial function using a graphing calculator. Identify the x-intercepts, local

max and min, intervals of increasing and decreasing, end behavior, and domain and range.

5. π’š = (𝒙 βˆ’ πŸ‘)(𝒙 + πŸ”) 6. π’š = βˆ’πŸ(𝒙 βˆ’ 𝟐)(𝒙 βˆ’ πŸ“)

7. π’š = (𝒙 + πŸ‘)𝟐(𝒙 βˆ’ 𝟏) 8. π’š = βˆ’πŸ’(𝒙 βˆ’ 𝟐)(𝒙 + πŸ’)πŸ‘(𝒙 + πŸ”)

9. π’š = (𝒙 + 𝟏)πŸπŸ’(𝒙 βˆ’ πŸ‘)𝟐 10. π’š = βˆ’π’™(𝒙 + πŸ‘)(𝒙 βˆ’ πŸ’)

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Graph each polynomial function by hand. Identify the x-intercepts and end behavior.

11. π’š = βˆ’πŸ”π’™(πŸ’π’™ βˆ’ πŸ‘)𝟐(πŸπ’™ + πŸ–)πŸ• 12. π’š = (πŸ”π’™ + 𝟏𝟐)(𝒙 βˆ’ πŸ‘)(πŸ‘π’™ + πŸπŸ“)πŸ‘

13. π’š = πŸπ’™πŸ’ + +πŸ–π’™πŸ‘ βˆ’ πŸ’πŸπ’™πŸ 14. π’š = βˆ’πŸ‘π’™πŸ βˆ’ πŸπŸπ’™πŸ βˆ’ πŸ‘π’™

15. π’š = π’™πŸ‘ + πŸ‘π’™πŸ βˆ’ πŸ—π’™ βˆ’ πŸπŸ• 16. π’š = βˆ’π’™πŸ“ βˆ’ π’™πŸ’ + πŸπ’™πŸ‘

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Unit 2.6 Dividing Polynomials Using Long Division and Synthetic Division

Student Learning Targets:

I can divide polynomials using long division.

I can find the remainder using the Remainder Theorem

I can apply the ideas of the Factor Theorem

I can divide polynomials using synthetic division.

Remainder Theorem:

If a polynomial f(x) is divided by x – k, then the remainder is r = f(k).

Factor Theorem:

A polynomial function f(x) has a factor x – k, if and only if f(k) = 0.

Notes:

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Assignment 2.6:

Simplify.

1. 4π‘₯𝑦2βˆ’2π‘₯𝑦+2π‘₯2𝑦

π‘₯𝑦 2. (3π‘Ž2𝑏 βˆ’ 6π‘Žπ‘ + 5π‘Žπ‘2) Γ· (π‘Žπ‘)

Simplify using long division.

3. (10π‘₯2 + 15π‘₯ + 20) Γ· (5π‘₯ + 5) 4. π‘₯4βˆ’3π‘₯3+6π‘₯2βˆ’3π‘₯+5

π‘₯2+1

5. (18π‘Ž2 + 6π‘Ž + 9) Γ· (3π‘Ž βˆ’ 2) 6. 27𝑦2+27π‘¦βˆ’30

9π‘¦βˆ’6

Use the Remainder Theorem to find the remainder when f(x) is divided by x – k.

7. 𝑓(π‘₯) = 2π‘₯2 βˆ’ 3π‘₯ + 1; π‘˜ = 2 8. 𝑓(π‘₯) = π‘₯4 βˆ’ 5; π‘˜ = 1

Use the Factor Theorem to determine whether the first polynomial is a factor of

the second polynomial.

9. (π‘₯ βˆ’ 3); π‘₯3 βˆ’ π‘₯2 βˆ’ π‘₯ βˆ’ 15 10. (π‘₯ βˆ’ 2); π‘₯3 + 3π‘₯ βˆ’ 4

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Simplify using synthetic division.

11. (𝑧4 βˆ’ 3𝑧3 + 2𝑧2 βˆ’ 4𝑧 + 4) Γ· (𝑧 βˆ’ 1) 12. 𝑦3+11𝑦2βˆ’10𝑦+6

𝑦+2

13. 2π‘₯4βˆ’5π‘₯3+7π‘₯2βˆ’3π‘₯+1

π‘₯βˆ’3 14. (5π‘₯4 βˆ’ 3π‘₯ + 1) Γ· (π‘₯ βˆ’ 4)

15. π‘₯4βˆ’3π‘₯2βˆ’18

π‘₯βˆ’2

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Unit 2.7: Extend Polynomial Identities to Complex Numbers

Student Learning Targets:

Know what the Linear Factorization Theorem is and how to use it.

Find complex zeros of a polynomial function.

Write a linear factorization of a polynomial function.

Notes:

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Assignment 2.7:

Write each polynomial in standard form.

1. 𝑓(π‘₯) = (π‘₯ + 1)(π‘₯ βˆ’ 3)(π‘₯ + 5) 2. 𝑓(π‘₯) = (2π‘₯ βˆ’ 1)(π‘₯ + 4𝑖)(π‘₯ βˆ’ 4𝑖)

State the number of zeros each function has without graphing the function.

3. 𝑓(π‘₯) = π‘₯5 + 4π‘₯3 βˆ’ 6π‘₯2 + 2π‘₯ + 1 4. 𝑓(π‘₯) = βˆ’3π‘₯7 + 8π‘₯6 βˆ’ 2π‘₯3 + 4π‘₯2

Use the rational zeros theorem to identify all the possible rational zeros of each function.

5. 𝑓(π‘₯) = π‘₯4 + 8π‘₯ + 32 6. 𝑓(π‘₯) = π‘₯3 + π‘₯2 + π‘₯ βˆ’ 28

7. 𝑓(π‘₯) = π‘₯7 βˆ’ 9π‘₯5 + 5π‘₯4 βˆ’ 3π‘₯2 + 7 8. 𝑓(π‘₯) = π‘₯12 + 12π‘₯7 + 4π‘₯3 + π‘₯ βˆ’ 30

Given a polynomial and one of its factors, find the remaining factors of the polynomial.

9. 𝑓(π‘₯) = π‘₯3 βˆ’ π‘₯2 βˆ’ 10π‘₯ βˆ’ 8; π‘₯ + 2 10. 𝑓(π‘₯) = 2π‘₯3 + 7π‘₯2 βˆ’ 53π‘₯ βˆ’ 28; π‘₯ βˆ’ 4

11. 𝑓(π‘₯) = 2π‘₯3 + 17π‘₯2 + 23π‘₯ βˆ’ 42; π‘₯ βˆ’ 1 12. 𝑓(π‘₯) = 6π‘₯3 βˆ’ 25π‘₯2 + 2π‘₯ + 8; 2π‘₯ + 1

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Find all of the zeros of each function. Then write each polynomial in factored form.

13. 𝑓(π‘₯) = π‘₯3 + 10π‘₯2 + 31π‘₯ + 30

14. 𝑓(π‘₯) = π‘₯4 βˆ’ π‘₯3 βˆ’ π‘₯2 βˆ’ π‘₯ βˆ’ 2

15. 𝑓(π‘₯) = π‘₯3 + 6π‘₯2 βˆ’ π‘₯ βˆ’ 30

16. 𝑓(π‘₯) = π‘₯3 + π‘₯2 + 4π‘₯ + 4

True or False. Justify your answer.

17. The polynomial 𝑓(π‘₯) = 4π‘₯3 + 7π‘₯ + 5 has three zeros.

18. The polynomial 𝑓(π‘₯) = π‘₯4 + 5π‘₯ βˆ’ 4 has a zero at π‘₯ = 2.

19. A polynomial with degree of 4 will cross the x-axis 4 times.