1. Splash Screen

2. Lesson Menu 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

3. Over Lesson 5–6 5-Minute Check 1 A.(2c)(4c 2 + cg + g 2 ) B.(2c – g)(4c 2 + 2cg + g 2 ) C.(c – g)(2c + g + g 2 ) D.prime Factor 8c 3 – g 3. If the polynomial is not factorable, write prime.

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

5. Over Lesson 5–6 5-Minute Check 3 A.(8x + y)(m + n)(m – n) B.(4x + y)(2x – y 2 )(m + n)(m – n) C.(2x + y)(4x 2 – 2xy + y 2 )(m + n)(m – n) D.prime Factor 8x 3 m 2 – 8x 3 n 2 + y 3 m 2 – y 3 n 2. If the polynomial is not factorable, write prime.

6. Over Lesson 5–6 5-Minute Check 4 Solve 16d 4 – 48d 2 + 32 = 0. A. B. C. D.

7. Over Lesson 5–6 5-Minute Check 5 A.16 B.8 C.–2 D.–4 Solve k 3 + 64 = 0.

8. Over Lesson 5–6 5-Minute Check 6 A.2 ft B.3 ft C.4 ft D.6 ft 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.

9. CCSS 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.

10. Then/Now 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.

11. Vocabulary synthetic substitution depressed polynomial

12. Concept

13. Example 1 Synthetic Substitution If f(x) = 2x 4 – 5x 2 + 8x – 7, find f(6). Method 1 Synthetic Substitution Answer: The remainder is 2453. Thus, by using synthetic substitution, f(6) = 2453. By the Remainder Theorem, f(6) should be the remainder when you divide the polynomial by x – 6. 21267410 2453 Notice that there is no x 3 term. A zero is placed in this position as a placeholder. 2 0–5 8 –7 1272402 2460

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

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

16. Example 2 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 3 + 4.03x 2 + 0.09x + 77.54, 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.

17. Example 2 A.616,230 students B.638,680 students C.646,720 students D.659,910 students 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 3 + 0.04x 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?

18. Concept

19. Example 3 Use the Factor Theorem Determine whether x – 3 is a factor of x 3 + 4x 2 – 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. 17601760 14–15–18 32118

20. Example 3 Use the Factor Theorem Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x 3 + 4x 2 – 15x –18 can be factored as (x – 3)(x 2 + 7x + 6). The polynomial x 2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored. x 2 + 7x + 6 = (x + 6)(x + 1)Factor the trinomial. Answer: So, x 3 + 4x 2 – 15x – 18 = (x – 3)(x + 6)(x + 1).

21. Example 3 Use the Factor Theorem CheckYou can see that the graph of the related function f(x) = x 3 + 4x 2 – 15x – 18 crosses the x-axis at 3, –6, and –1. Thus, f(x) = (x – 3)[x – (–6)][x – (–1)].

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

23. End of the Lesson