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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 11

Find the partial fraction decomposition for each rational expression. See Examples 1–4. x/(x2 + 4x - 5)

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1
Start by factoring the denominator of the rational expression. The denominator is \(x^2 + 4x - 5\). To factor it, find two numbers that multiply to \(-5\) and add to \(4\).
Rewrite the denominator as a product of two binomials: \(x^2 + 4x - 5 = (x + 5)(x - 1)\).
Set up the partial fraction decomposition form. Since the denominator factors into two distinct linear factors, express the fraction as \(\frac{x}{(x + 5)(x - 1)} = \frac{A}{x + 5} + \frac{B}{x - 1}\), where \(A\) and \(B\) are constants to be determined.
Multiply both sides of the equation by the common denominator \((x + 5)(x - 1)\) to clear the fractions: \(x = A(x - 1) + B(x + 5)\).
Expand the right side and collect like terms: \(x = A x - A + B x + 5 B = (A + B) x + (-A + 5 B)\). Then, equate the coefficients of corresponding powers of \(x\) on both sides to form a system of equations: For \(x\) terms, \$1 = A + B\(; for constants, \)0 = -A + 5 B\(. Solve this system to find \)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 simpler terms with linear or quadratic denominators.
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Factoring Quadratic Expressions

Factoring quadratic expressions means rewriting a quadratic polynomial as a product of two binomials. For example, x^2 + 4x - 5 factors into (x + 5)(x - 1). This step is essential to identify the denominators in partial fractions.
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Setting Up and Solving Equations for Coefficients

After expressing the rational function as a sum of partial fractions, you set up equations by equating numerators. Solving these equations for unknown coefficients allows you to find the values that complete the decomposition.
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