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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
All textbooksBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblem 29
Chapter 6, Problem 29

Write the partial fraction decomposition of each rational expression. 5x2 -6x+7/(x − 1) (x2 + 1)

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1
Identify the form of the denominator. Here, the denominator is \( (x - 1)(x^2 + 1) \), which consists of a linear factor \( (x - 1) \) and an irreducible quadratic factor \( (x^2 + 1) \).
Set up the partial fraction decomposition with unknown constants. For the linear factor, use a constant numerator \( A \), and for the quadratic factor, use a linear numerator \( Bx + C \). So, write: \[ \frac{5x^2 - 6x + 7}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} \]
Multiply both sides of the equation by the denominator \( (x - 1)(x^2 + 1) \) to clear the fractions: \[ 5x^2 - 6x + 7 = A(x^2 + 1) + (Bx + C)(x - 1) \]
Expand the right-hand side by distributing \( A \) and then expanding \( (Bx + C)(x - 1) \). This will give a polynomial expression in terms of \( x \): \[ A x^2 + A + Bx^2 - Bx + Cx - C \]
Group like terms (powers of \( x \)) on the right side and equate the coefficients of corresponding powers of \( x \) from both sides to form a system of equations. Solve this system to find the values of \( A \), \( B \), and \( C \).

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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 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 the Denominator

Factoring the denominator involves expressing it as a product of linear and/or irreducible quadratic factors. In this problem, the denominator is already factored as (x − 1)(x² + 1), which guides the form of the partial fractions to include terms over each factor.
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Form of Partial Fractions for Linear and Quadratic Factors

For linear factors like (x − 1), the partial fraction takes the form A/(x − 1). For irreducible quadratic factors like (x² + 1), the partial fraction is expressed as (Bx + C)/(x² + 1). This ensures the decomposition accounts for all possible numerators matching the degree of the factors.
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Related Practice
Textbook Question

Graph the solution set of each system of inequalities or indicate that the system has no solution.

{2x−5y≤103x−2y>6\(\begin{cases}\)2x - 5y \(\leq\) 10 \\3x - 2y > 6\(\end{cases}\)

Textbook Question

Graph the solution set of each system of inequalities or indicate that the system has no solution.

{3x+6y≤62x+y≤8\(\begin{cases}\)3x + 6y \(\leq\) 6 \\2x + y \(\leq\) 8\(\end{cases}\)

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Textbook Question

In Exercises 25–35, solve each system by the method of your choice. This is a piecewise function, refer to textbook problem.

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Textbook Question

In Exercises 29–42, solve each system by the method of your choice. {3x2+4y2=162x2−3y2=5\(\begin{cases}\)3x^2 + 4y^2 = 16 \\2x^2 - 3y^2 = 5\(\end{cases}\)

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Textbook Question

In Exercises 19–30, solve each system by the addition method. 3x = 4y + 1 3y = 1 - 4x

Textbook Question

Graph the solution set of each system of inequalities or indicate that the system has no solution.

{y>2x−3y<−x+6\(\begin{cases}\)y > 2x - 3 \(\y\) < -x + 6\(\end{cases}\)