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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 25
Chapter 6, Problem 25

Write the partial fraction decomposition of each rational expression. x2+2x+7/x(x − 1)2

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Identify the form of the denominator and write the general form of the partial fraction decomposition. Since the denominator is \(x(x - 1)^2\), the decomposition will include terms for the linear factor \(x\) and the repeated linear factor \((x - 1)^2\).
Set up the partial fraction decomposition as: \(\frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}\), where \(A\), \(B\), and \(C\) are constants to be determined.
Multiply both sides of the equation by the common denominator \(x(x - 1)^2\) to clear the fractions, resulting in: \(x^2 + 2x + 7 = A(x - 1)^2 + Bx(x - 1) + Cx\).
Expand the right-hand side by applying the distributive property and simplifying each term: expand \((x - 1)^2\), multiply terms, and combine like terms.
Equate the coefficients of corresponding powers of \(x\) on both sides of the equation to form a system of linear equations in \(A\), \(B\), and \(C\). 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 fractions into manageable parts.
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Decomposition of Functions

Factorization of Denominators

Understanding how to factor the denominator is essential for setting up partial fractions. In this problem, the denominator x(x − 1)² includes a linear factor and a repeated linear factor, which affects the form of the decomposition by requiring terms for each power of the repeated factor.
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Rationalizing Denominators

Setting Up and Solving Equations for Coefficients

After expressing the rational expression as a sum of partial fractions, you equate the numerators and solve for unknown coefficients. This involves multiplying both sides by the common denominator and comparing coefficients of corresponding powers of x to form a system of equations.
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Related Practice
Textbook Question

Solve each system in Exercises 25–26. {x+26−y+43+z2=0x+12+y−12−z4=92x−54+y+13+z−22=194\(\begin{cases}\[\frac{x + 2}{6}\) - \(\frac{y + 4}{3}\) + \(\frac{z}{2}\) = 0 \(\frac{x + 1}{2}\) + \(\frac{y - 1}{2}\) - \(\frac{z}{4}\) = \(\frac{9}{2}\) \(\frac{x - 5}{4}\) + \(\frac{y + 1}{3}\) + \(\frac{z - 2}{2}\) = \(\frac{19}{4}\]\end{cases}\)

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Textbook Question

Graph each inequality. y≥log2(x+1)

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Textbook Question

In Exercises 19–30, solve each system by the addition method. 4x + 3y = 15 2x - 5y = 1

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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 19–30, solve each system by the addition method. 3x - 4y = 11 2x + 3y = - 4

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Textbook Question

Solve each system in Exercises 25–26. {x+32−y−12+z+24=32x−52+y+13−z4=−256x−34−y+12+z−32=−52\(\begin{cases}\[\frac{x + 3}{2}\) - \(\frac{y - 1}{2}\) + \(\frac{z + 2}{4}\) = \(\frac{3}{2}\) \(\frac{x - 5}{2}\) + \(\frac{y + 1}{3}\) - \(\frac{z}{4}\) = - \(\frac{25}{6}\) \(\frac{x - 3}{4}\) - \(\frac{y + 1}{2}\) + \(\frac{z - 3}{2}\) = - \(\frac{5}{2}\]\end{cases}\)

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