Explanation Given data: The given expression is ( 1 − 2 ) 4 + ( 1 + 2 ) 4 . Formula used: Consider the mathematical expression of binomial theorem using Pascal’s triangle . ( a + b ) n = c 0 a n b 0 + c 1 a n − 1 b 1 + c 2 a n − 2 b 2 + ⋯ + c n − 1 a 1 b n − 1 + c n a 0 b n (1) Calculation: Consider the term ( 1 − 2 ) 4 in a given expression. Substitute 1 for a , − 2 for b , and 4 for n in equation (1) to obtain the required expansion for ( 1 − 2 ) 4 . ( 1 + ( − 2 ) ) 4 = { c 0 ( 1 ) 4 ( − 2 ) 0 + c 1 ( 1 ) 4 − 1 ( − 2 ) 1 + c 2 ( 1 ) 4 − 2 ( − 2 ) 2 + c 3 ( 1 ) 4 − 3 ( − 2 ) 3 + c 4 ( 1 ) 4 − 4 ( − 2 ) 4 } ( 1 − 2 ) 4 = { c 0 ( 1 ) 4 ( − 2 ) 0 + c 1 ( 1 ) 3 ( − 2 ) 1 + c 2 ( 1 ) 2 ( − 2 ) 2 + c 3 ( 1 ) 1 ( − 2 ) 3 + c 4 ( 1 ) 0 ( − 2 ) 4 } ( 1 − 2 ) 4 = c 0 − 2 c 1 + 2 c 2 − 2 2 c 3 + 4 c 4 (2) The coefficient values (from Pascal’s triangle) for n = 4 are, c 0 = 1 , c 1 = 4 , c 2 = 6 , c 3 = 4 , c 4 = 1 Substitute 1 for c 0 , 4 for c 1 , 6 for c 2 , 4 for c 3 , and 1 for c 4 in equation (2) to obtain the required value for ( 1 − 2 ) 4 . ( 1 − 2 ) 4 = 1 − 2 ( 4 ) + 2 ( 6 ) − 2 2 ( 4 ) + 4 ( 1 ) = 1 − 4 2 + 12 − 8 2 + 4 = 17 − 12 2 Consider the term ( 1 + 2 ) 4 in a given expression

Algebra and Trigonometry: Graphs and Models (6th Edition)

6th Edition

ISBN: 9780134179049

Author: Marvin L. Bittinger, Judith A. Beecher, David J. Ellenbogen, Judith A. Penna

Publisher: PEARSON

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Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

Chapter 11.7, Problem 18E

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Expand the expression (1−2)4+(1+2)4.

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Chapter 11.7, Problem 17E

Chapter 11.7, Problem 19E

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Chapter 11 Solutions

Algebra and Trigonometry: Graphs and Models (6th Edition)

Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

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Answers may vary. 58. 3 + 6...Ch. 11.1 - Prob. 59E Ch. 11.1 - Prob. 60E Ch. 11.1 - Prob. 61E Ch. 11.1 - Prob. 62E Ch. 11.1 - Prob. 63E Ch. 11.1 - Prob. 64E Ch. 11.1 - Prob. 65E Ch. 11.1 - Prob. 66E Ch. 11.1 - Prob. 67E Ch. 11.1 - Prob. 68E Ch. 11.1 - Prob. 69E Ch. 11.1 - Prob. 70E Ch. 11.1 - Prob. 71E Ch. 11.1 - Prob. 72E Ch. 11.1 - Prob. 73E Ch. 11.1 - Prob. 74E Ch. 11.1 - Prob. 75E Ch. 11.1 - Prob. 78E Ch. 11.1 - Prob. 79E Ch. 11.1 - Prob. 80E Ch. 11.1 - Prob. 81E Ch. 11.1 - Prob. 82E Ch. 11.1 - Prob. 83E Ch. 11.1 - Prob. 84E Ch. 11.1 - Prob. 85E Ch. 11.1 - Prob. 86E Ch. 11.2 - Prob. 1E Ch. 11.2 - Prob. 2E Ch. 11.2 - Prob. 3E Ch. 11.2 - Prob. 4E Ch. 11.2 - Prob. 5E Ch. 11.2 - Prob. 6E Ch. 11.2 - Prob. 7E Ch. 11.2 - Prob. 8E Ch. 11.2 - Prob. 9E Ch. 11.2 - Prob. 10E Ch. 11.2 - Prob. 11E Ch. 11.2 - Prob. 12E Ch. 11.2 - Prob. 13E Ch. 11.2 - Prob. 14E Ch. 11.2 - Prob. 15E Ch. 11.2 - Prob. 16E Ch. 11.2 - Prob. 17E Ch. 11.2 - Prob. 18E Ch. 11.2 - Prob. 19E Ch. 11.2 - Prob. 20E Ch. 11.2 - Prob. 21E Ch. 11.2 - Prob. 22E Ch. 11.2 - Prob. 23E Ch. 11.2 - Prob. 24E Ch. 11.2 - Prob. 25E Ch. 11.2 - Prob. 26E Ch. 11.2 - Prob. 27E Ch. 11.2 - Prob. 28E Ch. 11.2 - Prob. 29E Ch. 11.2 - 30. 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Prob. 52E Ch. 11.7 - Prob. 53E Ch. 11.7 - Find the sum. 54. 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Chapter 11 Solutions