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The Binomial Theorem

The Binomial Theorem. ALGEBRA 2 LESSON 6-8. (For help, go to Lessons 5-1 and 6-7.). Multiply. 1. ( x + 2) 2 2. (2 x + 3) 2 3. ( x – 3) 3 4. ( a + b ) 4 Evaluate. 5. 5 C 0 6. 5 C 1 7. 5 C 2 8. 5 C 3 9. 5 C 4. 6-8. Solutions.

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The Binomial Theorem

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  1. The Binomial Theorem ALGEBRA 2 LESSON 6-8 (For help, go to Lessons 5-1 and 6-7.) Multiply. 1. (x + 2)22. (2x + 3)23. (x – 3)34. (a + b)4 Evaluate. 5. 5C06.5C17.5C28.5C39.5C4 6-8

  2. Solutions 1. (x + 2)2 = x2 + 2(2)x + 22 = x2 + 4x + 4 2. (2x + 3)2 = (2x)2 + 2(3)(2x) + 32 = 4x2 + 12x + 9 3. (x – 3)3 = (x – 3)(x – 3)2 = (x – 3)(x2 – 2(3)x + 32) = (x – 3)(x2 – 6x + 9) = x(x2 – 6x + 9) – 3(x2 – 6x + 9) = x3 – 6x2 + 9x – 3x2 + 18x – 27 = x3 – 9x2 + 27x – 27 4. (a + b)4 = (a + b)2(a + b)2 = (a2 + 2ab + b2)(a2 + 2ab + b2) = a2(a2 + 2ab + b2) + 2ab(a2 + 2ab + b2) + b2(a2 + 2ab + b2) = a4 + 2a3b + a2b2 + 2a3b + 4a2b2 + 2ab3 + a2b2 + 2ab3 + b4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 The Binomial Theorem ALGEBRA 2 LESSON 6-8 6-8

  3. 1 1 / 5! 1 • 5! 5! 0! (5 – 0)! 5.5C0 = = = = 1 6.5C1 = = = = = 5 7.5C2 = = = = = 10 8.5C3 = = = = = 10 9.5C4 = = = = = 5 / 5 1 / / / / 5! 1! • 4! 5 • 4 • 3 • 2 • 1 1 • 4 • 3 • 2 • 1 5! 1! (5 – 1)! / / / / 20 2 / / / 5! 2! (5 – 2)! 5! 2! • 3! 5 • 4 • 3 • 2 • 1 2 • 1 • 3 • 2 • 1 / / / / / / 5 • 4 • 3 • 2 • 1 3 • 2 • 1 • 2 • 1 20 2 5! 3! (5 – 3)! 5! 3! • 2! / / / 5! 4! (5 – 4)! 5! 4! • 1! / / / / 5 1 5 • 4 • 3 • 2 • 1 4 • 3 • 2 • 1 • 1 / / / / The Binomial Theorem ALGEBRA 2 LESSON 6-8 Solutions (continued) 6-8

  4. Assignment 1 • Page 342 2-32 even, 55, 56, 68

  5. pages 342–345  Exercises 1. 120 2. 3,628,800 3. 6,227,020,800 4. 720 5. 665,280 6. 120 7. 120 8. 3003 9. a. 24 b. 120 10. 8 11. 56 12. 336 13. 1680 14. 6 15. 120 16. 60,480 17. 60 18. 10,897,286,400 19. 4,151,347,200 20. 12 21. 15 22. 56 23. 1 24. 4 25. 35 26. 15 27. 35 28. 29. 4368 30. 21 31. 2600 32. 126 33. true because of the Comm. Prop. of Add. 5 18 Permutations and Combinations ALGEBRA 2 LESSON 6-7 6-7

  6. 34. true because of the Assoc. Prop. of Mult. 35. False; answers may vary. Sample: (3 + 2)! = 120 and 3! + 2! = 8 36. False; answers may vary. Sample: (3  2)! = 720 and 3!  2! = 12 37. False; answers may vary. Sample: (3!)! = 720 and (3!)2 = 36 38. False; answers may vary. Sample: (3!)2 = 36 and 3(2!) = 9 39. 3125 40. 60 41. 2 ways, because order matters 42. 0.000206 43. 84 44. 45. 5 46. permutation 47. permutation 48. combination 49. combination 50. 330,791,175 51. 210 52. 12,650 53. 5 4 19,393 12 49 Permutations and Combinations ALGEBRA 2 LESSON 6-7 6-7

  7. Permutations and Combinations ALGEBRA 2 LESSON 6-7 54. 9504 55.a. 56 b. 56 c. Answers may vary. Sample: Each time you choose 3 of the 8 points to use as vertices of a , the 5 remaining points could be used to form a pentagon. 56. 120 57. 3024 58. 360 59. 24 60. 1680 61. 840 62. 5040 63. 0 64–67. Check students’ work. 68.a. 2048 b. Answers may vary. Sample: No, because there are too many possible solutions. 69. a. The graph for y = xCx–2 is identical to the graph for y = xC2 because 2C2–2 = 2C2, 3C3–2 = 3C2, 4C4–2 = 4C2, 5C5–2 = 5C2, etc. b. Answers may vary. Sample: The function is defined only at discrete whole-number values of x, and not over a smooth range of points as in a continuous function. 6-7

  8. The Binomial Theorem • Expand a binomial You expand a binomial by multiplying and writing the resulting polynomial in standard form. • Pascal's Triangle Pascal's Triangle is a pattern for finding the coefficients of the terms of a binomial expansion.

  9. The exponents for a begin with 5 and decrease. 1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5 The exponents for b begin with 0 and increase. The Binomial Theorem ALGEBRA 2 LESSON 6-8 Use Pascal’s Triangle to expand (a + b)5. Use the row that has 5 as its second number. In its simplest form, the expansion is a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5. 6-8

  10. 1 4 6 4 1   Coefficients from Pascal’s Triangle. (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 The Binomial Theorem ALGEBRA 2 LESSON 6-8 Use Pascal’s Triangle to expand (x – 3)4. First write the pattern for raising a binomial to the fourth power. Since (x – 3)4 = (x + (–3))4, substitute x for a and –3 for b. (x + (–3))4 = x4 + 4x3(–3) + 6x2(–3)2 + 4x(–3)3 + (–3)4 = x4 – 12x3 + 54x2 – 108x + 81 The expansion of (x – 3)4 is x4 – 12x3 + 54x2 – 108x + 81. 6-8

  11. Binomial Theorem For every positive integer n, (a + b)n = nC0an + nC1an − 1b1 + ... + nCn − 1a1bn − 1 + nCnbn.

  12. The Binomial Theorem ALGEBRA 2 LESSON 6-8 Use the Binomial Theorem to expand (x – y)9. Write the pattern for raising a binomial to the ninth power. (a + b)9 = 9C0a9 + 9C1a8b + 9C2a7b2 + 9C3a6b3 + 9C4a5b4 + 9C5a4b5 + 9C6a3b6 + 9C7a2b7 + 9C8ab8 + 9C9b9 Substitute x for a and –y for b. Evaluate each combination. (x – y)9 = 9C0x9 + 9C1x8(–y) + 9C2x7(–y)2 + 9C3x6(–y)3 + 9C4x5(–y)4 + 9C5x4(–y)5 + 9C6x3(–y)6 + 9C7x2(–y)7 + 9C8x(–y)8 + 9C9(–y)9 = x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9 The expansion of (x – y)9 is x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9. 6-8

  13. 12! 5! •7! = • (0.9)7(0.1)5The probability p of success = 90%, or 0.9. The Binomial Theorem ALGEBRA 2 LESSON 6-8 Dawn Staley makes about 90% of the free throws she attempts. Find the probability that Dawn makes exactly 7 out of 12 consecutive free throws. Since you want 7 successes (and 5 failures), use the term p7q5. This term has the coefficient 12C5. Probability (7 out of 10) = 12C5p7q5 = 0.0037881114 Simplify. Dawn Staley has about a 0.4% chance of making exactly 7 out of 12 consecutive free throws. 6-8

  14. Assignment 2 • Page 349 6-22 even, 42-44, 59

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