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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblema 16
Capitolo 6, Problema 16

Write the partial fraction decomposition of each rational expression. x/(x2 +2x -3)

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Start by factoring the denominator of the rational expression. The denominator is \(x^2 + 2x - 3\). To factor it, look for two numbers that multiply to \(-3\) and add to \(2\).
Rewrite the denominator as a product of two binomials: \(x^2 + 2x - 3 = (x + a)(x + b)\), where \(a\) and \(b\) are the numbers found in the previous step.
Set up the partial fraction decomposition. Since the denominator factors into two linear factors, the decomposition will be of the form \(\frac{x}{(x + a)(x + b)} = \frac{A}{x + a} + \frac{B}{x + b}\), where \(A\) and \(B\) are constants to be determined.
Multiply both sides of the equation by the denominator \((x + a)(x + b)\) to clear the fractions, resulting in \(x = A(x + b) + B(x + a)\).
Expand the right side and collect like terms. Then, equate the coefficients of corresponding powers of \(x\) on both sides to form a system of equations. Solve this system to find the values of \(A\) and \(B\).

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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 rational expression into a sum of fractions with simpler denominators, typically linear or irreducible quadratic factors.
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Factoring Quadratic Expressions

Factoring quadratic expressions involves rewriting a quadratic polynomial as a product of two binomials. For example, x^2 + 2x - 3 factors into (x + 3)(x - 1). This step is essential in partial fraction decomposition to identify the denominators of the simpler 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 an equation by equating numerators. Then, by substituting values or comparing coefficients of like terms, you solve for unknown constants. This process determines the exact form of the decomposition.
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