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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 31

Write the partial fraction decomposition of each rational expression. 5x2+6x+3/(x + 1)(x² + 2x + 2)

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1
Identify the form of the denominator. Here, the denominator is \( (x + 1)(x^2 + 2x + 2) \), which consists of a linear factor \( (x + 1) \) and an irreducible quadratic factor \( (x^2 + 2x + 2) \).
Set up the partial fraction decomposition with unknown constants. For the linear factor \( (x + 1) \), use a constant numerator \( A \). For the irreducible quadratic factor \( (x^2 + 2x + 2) \), use a linear numerator \( Bx + C \). So, write:
\[ \frac{5x^2 + 6x + 3}{(x + 1)(x^2 + 2x + 2)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 2x + 2} \]
Multiply both sides of the equation by the denominator \( (x + 1)(x^2 + 2x + 2) \) to clear the fractions:
\[ 5x^2 + 6x + 3 = A(x^2 + 2x + 2) + (Bx + C)(x + 1) \]
Expand the right-hand side and then collect like terms (powers of \( x \)) to form an equation where the coefficients of corresponding powers of \( x \) on both sides are equal. This will give a system of equations to solve for \( A \), \( B \), and \( C \).

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Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions with denominators that are factors of the original denominator. This technique simplifies integration and other algebraic operations by breaking down complex expressions into manageable parts.
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Factoring polynomials involves expressing a polynomial as a product of its factors. Recognizing linear factors like (x + 1) and irreducible quadratic factors like (x² + 2x + 2) is essential for setting up the correct form of partial fractions.
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