 Download Download Presentation Splash Screen

# Splash Screen

Download Presentation ## Splash Screen

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. Splash Screen

2. Five-Minute Check (over Lesson 2-3) Then/Now New Vocabulary Key Concept: Rational Zero Theorem Example 1: Leading Coefficient Equal to 1 Example 2: Leading Coefficient not Equal to 1 Example 3: Real-World Example: Solve a Polynomial Equation Key Concept: Upper and Lower Bound Tests Example 4: Use the Upper and Lower Bound Tests Key Concept: Descartes’ Rule of Signs Example 5: Use Descartes’ Rule of Signs Key Concept: Fundamental Theorem of Algebra Key Concept: Linear Factorization Theorem Example 6: Find a Polynomial Function Given Its Zeros Key Concept: Factoring Polynomial Functions Over the Reals Example 7: Factor and Find the Zeros of a Polynomial Function Example 8: Find the Zeros of a Polynomial When One is Known Lesson Menu

3. A. B. C. D. Divide (2x4 – 3x3 – 13x2 + 36x – 45) ÷ (2x – 5) using long division. 5–Minute Check 1

4. A. B. C. D. Divide (2x4 – 3x3 – 13x2 + 36x – 45) ÷ (2x – 5) using long division. 5–Minute Check 1

5. A. B.x2 + 4x + 3 C. D.x2 – 4x + 1 Find (x3 + 2x2 – 5x – 6) ÷ (x – 2) using synthetic division. 5–Minute Check 2

6. A. B.x2 + 4x + 3 C. D.x2 – 4x + 1 Find (x3 + 2x2 – 5x – 6) ÷ (x – 2) using synthetic division. 5–Minute Check 2

7. A. B. C. D. Find (x4 – x3 + 2x + 5) ÷ (x + 2) using synthetic division. 5–Minute Check 3

8. A. B. C. D. Find (x4 – x3 + 2x + 5) ÷ (x + 2) using synthetic division. 5–Minute Check 3

9. Use the Factor Theorem to determine if (x – 2) and (x + 1) are factors of f(x) = x4 – 2x3 – 9x2 + 32x – 28. Use the binomials that are factors to write a factored form of f(x). A.yes, no; f(x) = (x – 2)(x3 – 9x + 14) B.no, yes; f(x) = (x + 1)(x3 – 3x2 – 6x + 28) C.yes, no; f(x) = (x – 2)2(x – 7) D.yes, yes; f(x) = (x – 2)(x + 1)(x2 + x – 8) 5–Minute Check 4

10. Use the Factor Theorem to determine if (x – 2) and (x + 1) are factors of f(x) = x4 – 2x3 – 9x2 + 32x – 28. Use the binomials that are factors to write a factored form of f(x). A.yes, no; f(x) = (x – 2)(x3 – 9x + 14) B.no, yes; f(x) = (x + 1)(x3 – 3x2 – 6x + 28) C.yes, no; f(x) = (x – 2)2(x – 7) D.yes, yes; f(x) = (x – 2)(x + 1)(x2 + x – 8) 5–Minute Check 4

11. Find f(3) using synthetic substitution iff(x) = x6 – 5x5 + 6x4 + 4x3 + 12x2 + 8. A.–2422 B.8 C.80 D.224 5–Minute Check 5

12. Find f(3) using synthetic substitution iff(x) = x6 – 5x5 + 6x4 + 4x3 + 12x2 + 8. A.–2422 B.8 C.80 D.224 5–Minute Check 5

13. You learned that a polynomial function of degree n can have at most n real zeros. (Lesson 2-1) • Find real zeros of polynomial functions. • Find complex zeros of polynomial functions. Then/Now

14. Rational Zero Theorem • lower bound • upper bound • Descartes’ Rule of Signs • Fundamental Theorem of Algebra • Linear Factorization Theorem • Conjugate Root Theorem • complex conjugates • irreducible over the reals Vocabulary

15. Key Concept 1

16. Leading Coefficient Equal to 1 A. List all possible rational zeros of f(x) = x3–3x2 – 2x + 4. Then determine which, if any, are zeros. Step 1 Identify possible rational zeros. Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 4. Therefore, the possible rational zeros are ±1, ±2, and ±4. Example 1

17. Leading Coefficient Equal to 1 Step 2 Use direct substitution to test each possible zero. f(1) = (1)3 – 3(1)2 – 2(1) + 4 or 0 f(–1) = (–1)3 – 3(–1)2 – 2(–1) + 4 or 2 f(2) = (2)3 – 3(2)2 – 2(2) + 4 or –4 f(–2) = (–2)3 – 3(–2)2 – 2(–2) + 4 or –12 f(4) = (4)3 – 3(4)2 – 2(4) + 4 or 12 f(–4) = (–4)3 – 3(–4)2 – 2(–4) + 4 or –100 The only rational zero is 1. Answer: Example 1

18. Leading Coefficient Equal to 1 Step 2 Use direct substitution to test each possible zero. f(1) = (1)3 – 3(1)2 – 2(1) + 4 or 0 f(–1) = (–1)3 – 3(–1)2 – 2(–1) + 4 or 2 f(2) = (2)3 – 3(2)2 – 2(2) + 4 or –4 f(–2) = (–2)3 – 3(–2)2 – 2(–2) + 4 or –12 f(4) = (4)3 – 3(4)2 – 2(4) + 4 or 12 f(–4) = (–4)3 – 3(–4)2 – 2(–4) + 4 or –100 The only rational zero is 1. Answer: ±1, ± 2, ± 4; 1 Example 1

19. Leading Coefficient Equal to 1 B. List all possible rational zeros of f(x) = x3 – 2x – 1. Then determine which, if any, are zeros. Step 1 Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term –1. Therefore, the possible rational zeros of f are 1 and –1. Example 1

20. 1 1 0 –2 –1 1 1 –1 1 1 –1 –2 –1 1 0 –2 –1 –1 1 1 1 –1 –1 0 Leading Coefficient Equal to 1 Step 2 Begin by testing 1 and –1 using synthetic substitution. Example 1

21. Leading Coefficient Equal to 1 Because f(–1) = 0, you can conclude that x = –1 is a zero of f. Thus f(x) = (x + 1)(x2 – x – 1). Because the factor x2 – x – 1 yields no rational zeros, we can conclude that f has only one rational zero, x = –1. Answer: Example 1

22. Leading Coefficient Equal to 1 Because f(–1) = 0, you can conclude that x = –1 is a zero of f. Thus f(x) = (x + 1)(x2 – x – 1). Because the factor x2 – x – 1 yields no rational zeros, we can conclude that f has only one rational zero, x = –1. Answer:±1; 1 Example 1

23. A. B. C. D. List all possible rational zeros of f(x) = x4 – 12x2 – 15x – 4. Then determine which, if any, are zeros. Example 1

24. A. B. C. D. List all possible rational zeros of f(x) = x4 – 12x2 – 15x – 4. Then determine which, if any, are zeros. Example 1

25. Leading Coefficient not Equal to 1 List all possible rational zeros of f(x) = 2x3– 5x2 – 28x + 15. Then determine which, if any, are zeros. Step 1 The leading coefficient is 2 and the constant term is 15. Possible rational zeros: Example 2

26. Testing each subsequent possible zero on the depressed polynomial, you can determine that x = 5 and are also rational zeros. By the division algorithm, f(x) = (x + 3)(x – 5)(2x – 1) so the rational zeros are x = –3, x = 5, and . Check this result by graphing. Leading Coefficient not Equal to 1 Step 2 By synthetic substitution, you can determine that x = –3 is a rational zero. Example 2

29. A. B. C. D. List all possible rational zeros of f(x) = 4x3 – 20x2 + x – 5. Then determine which, if any, are zeros. Example 2

30. A. B. C. D. List all possible rational zeros of f(x) = 4x3 – 20x2 + x – 5. Then determine which, if any, are zeros. Example 2

31. Solve a Polynomial Equation WATER LEVEL The water level in a bucket sitting on a patio can be modeled by f(x) = x3+ 4x2 –2x + 7, where f(x) is the height of the water in millimeters and x is the time in days. On what day(s) will the water reach a height of 10 millimeters? Because f(x) represents the day when the water level, you need to solve f(x) = 10 to determine what day the water will reach a height of 10 millimeters. Example 3

32. Solve a Polynomial Equation f(x) = 10 Write the equation. x3 + 4x2 – 2x + 7 = 10 Substitute x3 + 4x2 – 2x + 7 for f(x). x3 + 4x2 – 2x – 3 = 0 Subtract 10 from each side. Apply the Rational Zeros Theorem to this new polynomial function, f(x) = x3 + 4x2 – 2x – 3. Step 1 Possible rational zeros: factors of –3 = ±1, ±3. Example 3

33. 1 1 4 –2 –3 1 5 3 1 5 3 0 Solve a Polynomial Equation Step 2 Because the number of the day cannot be negative, check each of the positive rational zeros using synthetic substitution. Doing so, you can determine that x = 1 is the only positive rational zero of f. Because x = 1 is a zero of f, x = 1 is a solution of f(x) = 0. So, it was day 1 when the water reached a height of 10 millimeters. Answer: Example 3

34. 1 1 4 –2 –3 1 5 3 1 5 3 0 Solve a Polynomial Equation Step 2 Because the number of the day cannot be negative, check each of the positive rational zeros using synthetic substitution. Doing so, you can determine that x = 1 is the only positive rational zero of f. Because x = 1 is a zero of f, x = 1 is a solution of f(x) = 0. So, it was day 1 when the water reached a height of 10 millimeters. Answer:day 1 Example 3

35. A. 4 seconds, 10 seconds B. 4 seconds C. 5 seconds, seconds D. 5 seconds PHYSICS The path of a ball is given by the function f(x) = –4.9x2 + 21.5x + 40, where x is the time in seconds and f(x) is the height above the ground in meters. After how many seconds will the ball reach a height of 25 meters? Example 3

36. A. 4 seconds, 10 seconds B. 4 seconds C. 5 seconds, seconds D. 5 seconds PHYSICS The path of a ball is given by the function f(x) = –4.9x2 + 21.5x + 40, where x is the time in seconds and f(x) is the height above the ground in meters. After how many seconds will the ball reach a height of 25 meters? Example 3

37. Key Concept 2

38. Use the Upper and Lower Bound Tests Determine an interval in which all real zeros of f(x) = x4– 4x3 – 11x2 – 4x – 12 must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros. Step 1 Graph f(x). From this graph, it appears that the real zeros of this function lie in the interval [–3, 7]. Example 4

39. –3 1 –4 –11 –4 –12 –3 21 –30 102 1 –7 10 –34 90 Values alternate signs in the last line, so –3 is a lower bound. 7 1 –4 –11 –4 –12 7 21 70 462 1 3 10 66 450 Values are all nonnegative in last line, so 7 is an upper bound. Use the Upper and Lower Bound Tests Step 2 Test a lower bound of c = –3 and an upper bound of c = 7. Example 4

40. Use the Upper and Lower Bound Tests Step 3 Use the Rational Zero Theorem. Possible rational zeros: Factors of 12 = ±1, ±2, ±3, ±4 , ±6, ±12 . Because the real zeros are in the interval [–3, 7], you can narrow this list to just –1, –2, –3, 1, 2, 3, 4, and 6. From the graph it appears that only –2 and 6 are reasonable. Example 4

41. 6 1 –4 –11 –4 –12 6 12 6 12 1 2 1 2 0 –2 1 2 1 2 –2 0 –2 1 0 1 0 Use the Upper and Lower Bound Tests Begin by testing 6. Now test –2 in the depressed polynomial. Example 4

42. Use the Upper and Lower Bound Tests By the division algorithm, f(x) = (x – 6)(x + 2)(x2 + 1). Notice that the factor x2 + 1 has no real zeros associated with it because x2 + 1 = 0 has no real solutions. So f has two real solutions that are both rational, x = –2 and x = 6. The graph of f(x) = x4 – 4x3 – 11x2 – 4x – 12 supports this conclusion. Answer: Example 4

43. Use the Upper and Lower Bound Tests By the division algorithm, f(x) = (x – 6)(x + 2)(x2 + 1). Notice that the factor x2 + 1 has no real zeros associated with it because x2 + 1 = 0 has no real solutions. So f has two real solutions that are both rational, x = –2 and x = 6. The graph of f(x) = x4 – 4x3 – 11x2 – 4x – 12 supports this conclusion. Answer:Upper and lower bounds may vary. Sample answer: [–3, 7]; With synthetic division, the values alternate signs when testing –3, and are all nonnegative when testing 7. So, –3 is a lower bound and 7 is an upper bound. The zeros are –2 and 6. Example 4

44. A. [0, 4]; 1,2 B. [–1, 2]; 1, C. [–3, 5]; 1, D. [–2, 1]; 1, Determine an interval in which all real zeros of f(x) = 2x4 – 5x3 – 13x2 + 26x – 10 must lie. Then find all the real zeros. Example 4

45. A. [0, 4]; 1,2 B. [–1, 2]; 1, C. [–3, 5]; 1, D. [–2, 1]; 1, Determine an interval in which all real zeros of f(x) = 2x4 – 5x3 – 13x2 + 26x – 10 must lie. Then find all the real zeros. Example 4

46. Key Concept 3

47. f(x) = x4 – 3x3 – 5x2 + 2x + 7 + to – – to + = x4 + 3x3 – 5x2 – 2x + 7 + to – – to + Use Descartes’ Rule of Signs Describe the possible real zeros of f(x) = x4– 3x3 – 5x2 + 2x + 7. Examine the variations of sign for f(x) and for f(–x). f(–x) = (–x)4 – 3(–x)3 – 5(–x)2 + 2(–x) + 7 Example 5

48. Use Descartes’ Rule of Signs The original function f(x) has two variations in sign, while f(–x) also has two variations in sign. By Descartes' Rule of Signs, you know that f(x) has either 2 or 0 positive real zeros and either 2 or 0 negative real zeros. Answer: Example 5

49. Use Descartes’ Rule of Signs The original function f(x) has two variations in sign, while f(–x) also has two variations in sign. By Descartes' Rule of Signs, you know that f(x) has either 2 or 0 positive real zeros and either 2 or 0 negative real zeros. Answer:2 or 0 positive real zeros, 2 or 0 negative real zeros Example 5

50. Describe the possible real zeros of g(x) = –x3 + 8x2 – 7x + 9. A. 3 or 1 positive real zeros, 1 negative real zero B. 3 or 1 positive real zeros, 0 negative real zeros C. 2 or 0 positive real zeros, 0 negative real zeros D. 2 or 0 positive real zeros, 1 negative real zero Example 5