Explanation Given Information: The expression is ( 1 x − x ) 6 . Formula used: The Binomial expression of a + b for any natural number n using Pascal’s triangle is, ( 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 Where c 0 , c 1 , c 2 , ⋯ , c n are from the ( n + 1 ) st row of the Pascal’s triangle. The Binomial expression of a + b by using factorial notation, when n is any natural number is, ( a + b ) n = ( n 0 ) a n + ( n 1 ) a n − 1 b + ( n 2 ) a n − 2 b 2 + ⋯ + ( n n ) b n Calculation: Consider the expression, ( 1 x − x ) 6 . Use the Binomial expression, ( 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 to expand the above term. Here a = 1 x , b = − x , n = 6

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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 14.4, Problem 41ES

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To calculate: The expanded value of (1x−x)6 by Binomial theorem using Pascal’s triangle and factorial notation.

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Chapter 14.4, Problem 40ES

Chapter 14.4, Problem 42ES

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Chapter 14 Solutions

EBK ELEMENTARY AND INTERMEDIATE ALGEBRA

Ch. 14.1 - Prob. 1YT Ch. 14.1 - Prob. 2YT Ch. 14.1 - Prob. 3YT Ch. 14.1 - Prob. 4YT Ch. 14.1 - Prob. 5YT Ch. 14.1 - Prob. 1ES Ch. 14.1 - Prob. 2ES Ch. 14.1 - Prob. 3ES Ch. 14.1 - Prob. 4ES Ch. 14.1 - Prob. 5ES

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The expanded value of ( 1 x − x ) 6 by Binomial theorem using Pascal’s triangle and factorial notation.

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To calculate: The expanded value of (1x−x)6 by Binomial theorem using Pascal’s triangle and factorial notation.

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Chapter 14 Solutions