1. Splash Screen

3. Five-Minute Check (over Lesson 2-2) Then/Now New Vocabulary
Example 1: Use Long Division to Factor Polynomials
Key Concept: Polynomial Division
Example 2: Long Division with Nonzero Remainder
Example 3: Division by Polynomial of Degree 2 or Higher
Key Concept: Synthetic Division Algorithm
Example 4: Synthetic Division
Key Concept: Remainder Theorem
Example 5: Real-World Example: Use the Remainder Theorem
Key Concept: Factor Theorem
Example 6: Use the Factor Theorem
Concept Summary: Synthetic Division and Remainders
Lesson Menu

4. Graph f (x) = (x – 2)3 + 3. A. B. C. D.
5–Minute Check 1

5. Falls left & Rises Right
Describe the end behavior of the graph of f (x) = 2x3 – 4x + 1 using limits. Explain your reasoning using the leading term test.
A. The degree is 3 and the leading coefficient 2. Because the degree is odd and the leading coefficient is positive,
B. The degree is 3 and the leading coefficient 2. Because the degree is odd and the leading coefficient is positive,
C. The degree is 3, and the leading coefficient is 2. Because the degree is odd and the leading coefficient is positive,
D. The degree is 3 and the leading coefficient is 2. Because the degree is odd and the leading coefficient is positive,
Falls left & Rises Right
5–Minute Check 2

6. Determine all of the real zeros of f (x) = 4x 6 – 16x 4.
B. 0, −2, 2, 4
C. −2, 2
D. −4, 4
5–Minute Check 3

7. You factored quadratic expressions to solve equations. (Lesson 0–3)
Divide polynomials using long division and synthetic division.
Use the Remainder and Factor Theorems.
Then/Now

8. synthetic substitution
synthetic division
depressed polynomial
synthetic substitution
Vocabulary

9. THE REMAINDER & FACTOR THEOREMS
Long Division
DAY 1
HW: Pg. 115, #’s 1-18

10. Use Long Division to Factor Polynomials
Factor 6x x 2 – 104x + 60 completely using long division if (2x – 5) is a factor.
←Multiply divisor by 3x 2 because = 3x 2.
(–)6x 3 – 15x 2
32x 2 – 104x
←Subtract and bring down next term.
←Multiply divisor by 16x because = 16x.
(–)32x 2 – 80x
←Subtract and bring down next term.
–24x + 60
←Multiply divisor by –12 because = –12
(–)–24x + 60
←Subtract. Notice that the remainder is 0.
Example 1

11. Use Long Division to Factor Polynomials
From this division, you can write 6x3 + 17x2 – 104x + 60 = (2x – 5)(3x2 + 16x – 12).
Factoring the quadratic expression yields 6x3 + 17x2 – 104x + 60 = (2x – 5)(3x – 2)(x + 6).
Answer: (2x – 5)(3x – 2)(x + 6)
Example 1

12. Factor 6x 3 + x 2 – 117x + 140 completely using long division if (3x – 4) is a factor.
A. (3x – 4)(x – 5)(2x + 7)
B. (3x – 4)(x + 5)(2x – 7)
C. (3x – 4)(2x 2 + 3x – 35)
D. (3x – 4)(2x + 5)(x – 7)
Example 1

13. Key Concept 1

14. Divide 6x 3 – 5x 2 + 9x + 6 by 2x – 1. (–)6x 3 – 3x 2 –2x 2 + 9x
Long Division with Nonzero Remainder
Divide 6x 3 – 5x 2 + 9x + 6 by 2x – 1.
(–)6x 3 – 3x 2
–2x 2 + 9x
(–)–2x x
8x + 6
(–)8x – 4 10
Example 2

15. You can write the result as .
Long Division with Nonzero Remainder
You can write the result as
Answer:
Check Multiply to check this result.
(2x – 1)(3x2 – x + 4) + 10 = 6x3 – 5x2 + 9x + 6
6x3 – 2x2 + 8x – 3x2 + x – = 6x3 – 5x2 + 9x + 6
6x3 – 5x2 + 9x + 6 = 6x3 – 5x2 + 9x + 6 
Example 2

16. Divide 4x 4 – 2x 3 + 8x – 10 by x + 1. A. B. 4x 3 + 2x 2 + 2x + 10 C.
Example 2

17. Divide x 3 – x 2 – 14x + 4 by x 2 – 5x + 6. (–)x 3 – 5x 2 + 6x
Division by Polynomial of Degree 2 or Higher
Divide x 3 – x 2 – 14x + 4 by x 2 – 5x + 6.
(–)x 3 – 5x 2 + 6x
4x 2 – 20x + 4
(–)4x 2 – 20x + 24
–20
Example 3

18. You can write this result as .
Division by Polynomial of Degree 2 or Higher
You can write this result as
Answer:
Example 3

19. Divide 2x 4 + 9x 3 + x2 – x + 26 by x 2 + 6x + 9.A. B. C. D.
Example 3

20. THE REMAINDER & FACTOR THEOREMS
Synthetic Division
LESSON 2
HW: Pg. 115, #’s 19-28

21. Key Concept 2

22. Factor 6x 3 + 17x 2 – 104x + 60 completely using Synthetic division if (2x – 5) is a factor.

23. = Multiply by c, and write the product.
Synthetic Division
A. Find (2x 5 – 4x 4 – 3x 3 – 6x 2 – 5x – 8) ÷ (x – 3) using synthetic division.
Because x – 3 is x – (3), c = 3. Set up the synthetic division as follows. Then follow the synthetic division procedure.
–4 –3 –6 –5 –8
= add terms. 6 6 9 9 12 2 2 3 3 4 4
= Multiply by c, and write the product.
coefficients of depressed quotient
remainder
Example 4

24. The quotient has degree one less than that of its dividend, so
Synthetic Division
The quotient has degree one less than that of its dividend, so
Answer:
Example 4

25. Synthetic Division
B. Find (8x x 3 + 5x 2 + 3x + 3) ÷ (4x + 1) using synthetic division.
Rewrite the division expression so that the divisor is of the form x – c.
Example 4

26. So, . Perform the synthetic division.
Example 4

27. Synthetic Division
So,
Answer:
Example 4

28. Find (6x 4 – 2x 3 + 8x 2 – 9x – 3) ÷ (x – 1) using synthetic division.A. B.
C. 6x3 – 8x2 + 3
D. 6x3 + 4x2 + 12x + 3
Example 4

29. THE REMAINDER & FACTOR THEOREMS Remainder Theorem

30. If f(x)= (6x 4 – 2x 3 + 8x 2 – 9x – 3) and is divided by (x – 1) , then f(1) will be the remainder.
Key Concept 3

32. Use the Remainder Theorem
REAL ESTATE Suppose 800 units of beachfront property have tenants paying $600 per week. Research indicates that for each $10 decrease in rent, 15 more units would be rented. The weekly revenue from the rentals is given by R (x) = –150x x + 480,000, where x is the number of $10 decreases the property manager is willing to take. Use the Remainder Theorem to find the revenue from the properties if the property manager decreases the rent by $50.
To find the revenue from the properties, use synthetic substitution to evaluate f (x) for x = 5 since $50 is 5 times $10.
Example 5

33. Use the Remainder Theorem
5 – ,000 –
– ,250
The remainder is 481,250, so f (5) = 481,250. Therefore, the revenue will be $481,250 when the rent is decreased by $50.
Example 5

34. Check You can check your answer using direct substitution.
Use the Remainder Theorem
Answer: $481,250
Check You can check your answer using direct substitution.
R(x) = –150x x + 480, Original function
R(5) = –150(5) (5) + 480,000 Substitute 5 for x.
R(5) = – ,000 or 481,250 Simplify.
Example 5

35. REAL ESTATE Use the equation for R(x) from Example 5 and the Remainder Theorem to find the revenue from the properties if the property manager decreases the rent by $100.
A. $380,000
B. $450,000
C. $475,000
D. $479,900
Example 5

36. THE REMAINDER & FACTOR THEOREMS Factor Theorem

37. Key Concept 4

38. Use synthetic division to test each factor, (x – 5) and (x + 5).
Use the Factor Theorem
A. Use the Factor Theorem to determine if (x – 5) and (x + 5) are factors of f (x) = x 3 – 18x x Use the binomials that are factors to write a factored form of f (x).
Use synthetic division to test each factor, (x – 5) and (x + 5).
5 1 –
5 –65 –25
1 –13 –5 0
Example 6

39. Answer: f (x) = (x – 5)(x 2 – 13x – 5)
Use the Factor Theorem
–5 1 –
–5 115 –875
1 – –850
Because the remainder when f (x) is divided by (x – 5) is 0, f(5) = 0, and (x – 5) is a factor.
Because the remainder when f (x) is divided by (x + 5) is –850, f (–5) = –850 and (x + 5) is not a factor.
Because (x – 5) is a factor of f (x), we can use the quotient of f (x) ÷ (x – 5) to write a factored form of f(x).
Answer: f (x) = (x – 5)(x 2 – 13x – 5)
Example 6

40. Use synthetic division to test the factor (x – 5).
Use the Factor Theorem
B. Use the Factor Theorem to determine if (x – 5) and (x + 2) are factors of f (x) = x 3 – 2x 2 – 13x – 10. Use the binomials that are factors to write a factored form of f (x).
Use synthetic division to test the factor (x – 5).
5 1 –2 –13 –10
Because the remainder when f (x) is divided by (x – 5) is 0, f (5) = 0 and (x – 5) is a factor of f (x).
Example 6

41. Answer: f (x) = (x – 5)(x + 2)(x + 1)
Use the Factor Theorem
Next, test the second factor (x + 2), with the depressed polynomial x2 + 3x + 2. –
–2 –2
1 1 0
Because the remainder when the quotient of f (x) ÷ (x – 5) is divided by (x + 2) is 0, f(–2) = 0 and (x + 2) is a factor of f (x).
Because (x – 5) and (x + 2) are factors of f (x), we can use the final quotient to write a factored form of f (x).
Answer: f (x) = (x – 5)(x + 2)(x + 1)
Example 6

42. A. yes, yes; f(x) = (x + 2)(x – 3)(–14x + 5)
Use the Factor Theorem to determine if the binomials (x + 2) and (x – 3) are factors of f (x) = 4x 3 – 9x 2 – 19x Use the binomials that are factors to write a factored form of f (x).
A. yes, yes; f(x) = (x + 2)(x – 3)(–14x + 5)
B. yes, yes; f(x) = (x + 2)(x – 3)(4x – 5)
C. yes, no; f(x) = (x + 2)(4x2 – 17x – 15)
D. no, yes; f(x) = (x – 3)(4x2 + 3x + 10)
Example 6

43. Key Concept 5

44. End of the Lesson