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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 6, Problem 21

Write the partial fraction decomposition of each rational expression. (6x-11)/(x − 1)²

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1
Identify the form of the denominator. Since the denominator is \( (x - 1)^2 \), which is a repeated linear factor, the partial fraction decomposition will include terms for each power of \( (x - 1) \) up to 2.
Set up the partial fraction decomposition as \( \frac{6x - 11}{(x - 1)^2} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} \), where \( A \) and \( B \) are constants to be determined.
Multiply both sides of the equation by the denominator \( (x - 1)^2 \) to clear the fractions: \( 6x - 11 = A(x - 1) + B \).
Expand the right side: \( 6x - 11 = A x - A + B \).
Group like terms and equate coefficients of corresponding powers of \( x \) on both sides to form a system of equations to solve for \( A \) and \( B \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions, making integration or other operations easier. It involves breaking down a complex fraction into a sum of fractions with simpler denominators, typically linear or quadratic factors.
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Repeated Linear Factors

When the denominator contains repeated linear factors, such as (x - 1)², the decomposition includes terms for each power of the factor up to its multiplicity. For (x - 1)², the decomposition will have terms with denominators (x - 1) and (x - 1)².
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Setting Up and Solving Equations for Coefficients

To find the unknown coefficients in the partial fractions, multiply both sides by the common denominator and equate coefficients of corresponding powers of x. This results in a system of equations that can be solved to determine the values of the constants.
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