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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 27
Chapter 6, Problem 27

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

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Identify the denominator and its factors. The denominator is \((x - 1)^2 (x + 1)\), which consists of a repeated linear factor \((x - 1)^2\) and a distinct linear factor \((x + 1)\).
Set up the form of the partial fraction decomposition. For the repeated linear factor \((x - 1)^2\), include terms with denominators \((x - 1)\) and \((x - 1)^2\). For the linear factor \((x + 1)\), include a term with denominator \((x + 1)\). So, write the decomposition as: \[\frac{x^2}{(x - 1)^2 (x + 1)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 1}\]
Multiply both sides of the equation by the common denominator \((x - 1)^2 (x + 1)\) to clear the denominators. This gives: \[x^2 = A(x - 1)(x + 1) + B(x + 1) + C(x - 1)^2\]
Expand the right-hand side by multiplying out each term: - Expand \(A(x - 1)(x + 1)\) using the difference of squares formula. - Expand \(B(x + 1)\). - Expand \(C(x - 1)^2\) by squaring the binomial.
Collect like terms on the right-hand side and equate the coefficients of corresponding powers of \(x\) from both sides. This will give a system of equations to solve for \(A\), \(B\), 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 complex rational expression as a sum of simpler fractions. This technique is especially useful for integrating rational functions or solving equations. It involves breaking down a fraction into components with denominators that are factors of the original denominator.
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Decomposition of Functions

Repeated Linear Factors

When the denominator contains repeated linear factors, such as (x - 1)², the decomposition must include terms for each power of the repeated factor. For example, for (x - 1)², the decomposition includes terms with denominators (x - 1) and (x - 1)², each with its own constant numerator.
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Setting Up and Solving Equations for Coefficients

After expressing the rational function as a sum of partial fractions, you multiply both sides by the common denominator to clear fractions. Then, by equating coefficients of corresponding powers of x or substituting convenient values, you solve for the unknown constants in the numerators.
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Related Practice
Textbook Question

In Exercises 19–28, solve each system by the addition method. {y2−x=4x2+y2=4\(\begin{cases}\)y^2 - x = 4 \(\x\)^2 + y^2 = 4\(\end{cases}\)

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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 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

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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