Explanation Given: The rational expression is 2 x − 6 ( x − 1 ) ( x − 2 ) 2 Calculation: As the linear factor x − 2 occurs multiple times in the denominator of the rational expression, so, for each power of x − 2 assign an undefined constant Factor the denominator by the method of grouping as shown below: 2 x − 6 ( x − 1 ) ( x − 2 ) 2 = A ( x − 1 ) + B x − 2 + C ( x − 2 ) 2 Multiply both sides of the equation by ( x − 1 ) ( x − 2 ) 2 : ( x − 1 ) ( x − 2 ) 2 ⋅ 2 x − 6 ( x − 1 ) ( x − 2 ) 2 = ( x − 1 ) ( x − 2 ) 2 ( A x − 1 + B x − 2 + C ( x − 2 ) 2 ) ( x − 1 ) ( x − 2 ) 2 ⋅ 2 x − 6 ( x − 1 ) ( x − 2 ) 2 = { ( x − 1 ) ( x − 2 ) 2 ⋅ A x − 1 + ( x − 1 ) ( x − 2 ) 2 ⋅ B x − 2 + ( x − 1 ) ( x − 2 ) 2 ⋅ C ( x − 2 ) 2 } Divide out the common factors: ( x − 1 ) ( x − 2 ) 2 ⋅ 2 x − 6 ( x − 1 ) ( x − 2 ) 2 = { ( x − 1 ) ( x − 2 ) 2 ⋅ A x − 1 + ( x − 1 ) ( x − 2 ) ( x − 2 ) ⋅ B x − 2 + ( x − 1 ) ( x − 2 ) 2 ⋅ C ( x − 2 ) 2 } 2 x − 6 = ( x − 2 ) 2 ⋅ A + ( x − 1 ) ( x − 2 ) ⋅ B + ( x − 1 ) C 2 x −<

5th Edition

ISBN: 9780321837240

Author: Robert F. Blitzer

Publisher: PEARSON

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Chapter 8, Problem 20RE

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To calculate: The partial fraction decomposition of the rational expression

2x−6(x−1)(x−2)2

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Chapter 8, Problem 19RE

Chapter 8, Problem 21RE

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The partial fraction decomposition of the rational expression 2 x − 6 ( x − 1 ) ( x − 2 ) 2

Solution Summary: The author explains that the linear factor x-2 occurs multiple times in the denominator of the rational expression.

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To calculate: The partial fraction decomposition of the rational expression

2x−6(x−1)(x−2)2

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Chapter 8 Solutions

The partial fraction decomposition of the rational expression 2 x − 6 ( x − 1 ) ( x − 2 ) 2

Solution Summary: The author explains that the linear factor x-2 occurs multiple times in the denominator of the rational expression.

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To calculate: The partial fraction decomposition of the rational expression 2x−6(x−1)(x−2)2

Want to see the full answer?

Chapter 8 Solutions

To calculate: The partial fraction decomposition of the rational expression

2x−6(x−1)(x−2)2