1. Splash Screen

2. Five-Minute Check (over Lesson 5–5) CCSS Then/Now New Vocabulary
Key Concept: Remainder Theorem
Example 1: Synthetic Substitution
Example 2: Real-World Example: Find Function Values
Key Concept: Factor Theorem
Example 3: Use the Factor Theorem
Lesson Menu

3. Factor 8c3 – g3. If the polynomial is not factorable, write prime.
A. (2c)(4c2 + cg + g2)
B. (2c – g)(4c2 + 2cg + g2)
C. (c – g)(2c + g + g2)
D. prime
5-Minute Check 1

4. Factor 8c3 – g3. If the polynomial is not factorable, write prime.
A. (2c)(4c2 + cg + g2)
B. (2c – g)(4c2 + 2cg + g2)
C. (c – g)(2c + g + g2)
D. prime
5-Minute Check 1

5. Factor 12az – 6bz – 6cz + 10ax – 5bx – 5cx
Factor 12az – 6bz – 6cz + 10ax – 5bx – 5cx. If the polynomial is not factorable, write prime.
A. (2a – 3z)(5x – c)
B. (2a – 6z)(2a + b + c)
C. (5x + 6z)(2a – b – c)
D. prime
5-Minute Check 2

6. Factor 12az – 6bz – 6cz + 10ax – 5bx – 5cx
Factor 12az – 6bz – 6cz + 10ax – 5bx – 5cx. If the polynomial is not factorable, write prime.
A. (2a – 3z)(5x – c)
B. (2a – 6z)(2a + b + c)
C. (5x + 6z)(2a – b – c)
D. prime
5-Minute Check 2

7. Factor 8x3m2 – 8x3n2 + y3m2 – y3n2. If the polynomial is not factorable, write prime.
A. (8x + y)(m + n)(m – n)
B. (4x + y)(2x – y2)(m + n)(m – n)
C. (2x + y)(4x2 – 2xy + y2)(m + n)(m – n)
D. prime
5-Minute Check 3

8. Factor 8x3m2 – 8x3n2 + y3m2 – y3n2. If the polynomial is not factorable, write prime.
A. (8x + y)(m + n)(m – n)
B. (4x + y)(2x – y2)(m + n)(m – n)
C. (2x + y)(4x2 – 2xy + y2)(m + n)(m – n)
D. prime
5-Minute Check 3

9. Solve 16d4 – 48d = 0. A. B. C. D.
5-Minute Check 4

10. Solve 16d4 – 48d = 0. A. B. C. D.
5-Minute Check 4

11. Solve k = 0.
A. 16
B. 8
C. –2
D. –4
5-Minute Check 5

12. Solve k = 0.
A. 16
B. 8
C. –2
D. –4
5-Minute Check 5

13. The width of a box is 3 feet less than the length
The width of a box is 3 feet less than the length. The height is 4 feet less than the length. The volume of the box is 36 cubic feet. Find the length of the box.
A. 2 ft
B. 3 ft
C. 4 ft
D. 6 ft
5-Minute Check 6

14. The width of a box is 3 feet less than the length
The width of a box is 3 feet less than the length. The height is 4 feet less than the length. The volume of the box is 36 cubic feet. Find the length of the box.
A. 2 ft
B. 3 ft
C. 4 ft
D. 6 ft
5-Minute Check 6

15. Mathematical Practices 7 Look for and make use of structure.
Content Standards
A.APR.2 Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x).
F.IF.7.c Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior.
Mathematical Practices
7 Look for and make use of structure.
CCSS

16. Evaluate functions by using synthetic substitution.
You used the Distributive Property and factoring to simplify algebraic expressions.
Evaluate functions by using synthetic substitution.
Determine whether a binomial is a factor of a polynomial by using synthetic substitution.
Then/Now

17. synthetic substitution depressed polynomial
Vocabulary

18. Concept

19. If f(x) = 2x4 – 5x2 + 8x – 7, find f(6).
Synthetic Substitution
If f(x) = 2x4 – 5x2 + 8x – 7, find f(6).
Method 1 Synthetic Substitution
By the Remainder Theorem, f(6) should be the remainder when you divide the polynomial by x – 6.
Notice that there is no x3 term. A zero is placed in this position as a placeholder.
2 0 – –7
Answer:
Example 1

20. If f(x) = 2x4 – 5x2 + 8x – 7, find f(6).
Synthetic Substitution
If f(x) = 2x4 – 5x2 + 8x – 7, find f(6).
Method 1 Synthetic Substitution
By the Remainder Theorem, f(6) should be the remainder when you divide the polynomial by x – 6.
Notice that there is no x3 term. A zero is placed in this position as a placeholder.
2 0 – –7
Answer: The remainder is Thus, by using synthetic substitution, f(6) = 2453.
Example 1

21. Method 2 Direct Substitution
Synthetic Substitution
Method 2 Direct Substitution
Replace x with 6.
Original function
Replace x with 6.
Simplify.
Answer:
Example 1

22. Method 2 Direct Substitution
Synthetic Substitution
Method 2 Direct Substitution
Replace x with 6.
Original function
Replace x with 6.
Simplify.
Answer: By using direct substitution, f(6) = 2453.
Example 1

23. If f(x) = 2x3 – 3x2 + 7, find f(3).
A. 20
B. 34
C. 88
D. 142
Example 1

24. If f(x) = 2x3 – 3x2 + 7, find f(3).
A. 20
B. 34
C. 88
D. 142
Example 1

25. Find Function Values
COLLEGE The number of college students from the United States who study abroad can be modeled by the function S(x) = 0.02x 4 – 0.52x x x , where x is the number of years since 1993 and S(x) is the number of students in thousands. How many U.S. college students will study abroad in 2011?
Answer:
Example 2

26. Find Function Values
COLLEGE The number of college students from the United States who study abroad can be modeled by the function S(x) = 0.02x 4 – 0.52x x x , where x is the number of years since 1993 and S(x) is the number of students in thousands. How many U.S. college students will study abroad in 2011?
Answer: In 2011, there will be about 451,760 U.S. college students studying abroad.
Example 2

27. HIGH SCHOOL The number of high school students in the United States who hosted foreign exchange students can be modeled by the function F(x) = 0.02x 4 – 0.05x x 2 – 0.02x, where x is the number of years since 1999 and F(x) is the number of students in thousands. How many U.S. students will host foreign exchange students in 2013?
A. 616,230 students
B. 638,680 students
C. 646,720 students
D. 659,910 students
Example 2

28. HIGH SCHOOL The number of high school students in the United States who hosted foreign exchange students can be modeled by the function F(x) = 0.02x 4 – 0.05x x 2 – 0.02x, where x is the number of years since 1999 and F(x) is the number of students in thousands. How many U.S. students will host foreign exchange students in 2013?
A. 616,230 students
B. 638,680 students
C. 646,720 students
D. 659,910 students
Example 2

29. Concept

30. Use the Factor Theorem
Determine whether x – 3 is a factor of x3 + 4x2 – 15x – 18. Then find the remaining factors of the polynomial.
The binomial x – 3 is a factor of the polynomial if 3 is a zero of the related polynomial function. Use the factor theorem and synthetic division.
1 4 –15 –18
Example 3

31. x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial.
Use the Factor Theorem
Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6). The polynomial x2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored.
x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial.
Answer:
Example 3

32. x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial.
Use the Factor Theorem
Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6). The polynomial x2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored.
x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial.
Answer: So, x3 + 4x2 – 15x – 18 = (x – 3)(x + 6)(x + 1).
Example 3

33. Use the Factor Theorem
Check You can see that the graph of the related function f(x) = x3 + 4x2 – 15x – 18 crosses the x-axis at 3, –6, and –1. Thus, f(x) = (x – 3)[x – (–6)][x – (–1)]. 
Example 3

34. Determine whether x + 2 is a factor of x3 + 8x2 + 17x + 10
Determine whether x + 2 is a factor of x3 + 8x2 + 17x If so, find the remaining factors of the polynomial.
A. yes; (x + 5)(x + 1)
B. yes; (x + 5)
C. yes; (x + 2)(x + 3)
D. x + 2 is not a factor.
Example 3

35. Determine whether x + 2 is a factor of x3 + 8x2 + 17x + 10
Determine whether x + 2 is a factor of x3 + 8x2 + 17x If so, find the remaining factors of the polynomial.
A. yes; (x + 5)(x + 1)
B. yes; (x + 5)
C. yes; (x + 2)(x + 3)
D. x + 2 is not a factor.
Example 3

36. End of the Lesson