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Ch. 5 - Systems and Matrices
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 6, Problem 9

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (4x + 2)/((x + 2)(2x - 1))

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1
Identify the form of the partial fraction decomposition. Since the denominator is factored into linear factors \((x + 2)\) and \((2x - 1)\), set up the decomposition as \(\frac{4x + 2}{(x + 2)(2x - 1)} = \frac{A}{x + 2} + \frac{B}{2x - 1}\), where \(A\) and \(B\) are constants to be determined.
Multiply both sides of the equation by the common denominator \((x + 2)(2x - 1)\) to clear the fractions. This gives \(4x + 2 = A(2x - 1) + B(x + 2)\).
Expand the right-hand side: \(A(2x - 1) = 2Ax - A\) and \(B(x + 2) = Bx + 2B\). So the equation becomes \(4x + 2 = 2Ax - A + Bx + 2B\).
Group like terms on the right-hand side: \(4x + 2 = (2A + B)x + (-A + 2B)\).
Set up a system of equations by equating the coefficients of corresponding terms from both sides: For the \(x\) terms, \(4 = 2A + B\); for the constant terms, \(2 = -A + 2B\). Solve this system to find the values of \(A\) 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 with linear or quadratic denominators. This technique simplifies integration and other algebraic operations by breaking down complex fractions into manageable parts.
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Factoring Denominators

Factoring the denominator into linear or irreducible quadratic factors is essential before performing partial fraction decomposition. It allows the rational expression to be split into simpler fractions, each corresponding to a factor in the denominator.
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Rationalizing Denominators

Setting Up and Solving Equations for Coefficients

After expressing the rational function as a sum of unknown coefficients over the factored denominators, you multiply through by the common denominator and equate coefficients of like terms. Solving the resulting system of equations finds the values of these unknowns.
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