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Find the partial fraction decomposition for each rational expression. $$\frac{x+2}{(x+1)(x-1)}$$

Short Answer

Expert verified
The partial fraction decomposition is \(\frac{-1}{2(x+1)} + \frac{3}{2(x-1)}\).

Step by step solution

01

Identify the form of the partial fraction decomposition

In this expression, the denominator \((x+1)(x-1)\) consists of two distinct linear factors. Thus, the partial fraction decomposition will be of the form \(\frac{A}{x+1} + \frac{B}{x-1}\).
02

Multiply through by the common denominator

Multiply both sides by the common denominator \((x+1)(x-1)\) to eliminate the fractions. This gives: \[x + 2 = A(x-1) + B(x+1)\]
03

Expand the right side

Distribute the terms on the right side:\[x + 2 = Ax - A + Bx + B\] This simplifies to: \[x + 2 = (A + B)x + (B - A)\]
04

Compare coefficients

Equate the coefficients from both sides of the equation to find A and B. From \(x\) terms: \[A + B = 1\]From constant terms: \[B - A = 2\]
05

Solve the system of equations

We have the system of equations: 1. \(A + B = 1\) 2. \(B - A = 2\) Add these two equations to eliminate \(A\): \[A + B + B - A = 1 + 2 \]This simplifies to \(2B = 3\), so \(B = \frac{3}{2}\). Substitute back into the first equation: \[A + \frac{3}{2} = 1\]Thus, \(A = 1 - \frac{3}{2} = -\frac{1}{2}\).
06

Write the partial fraction decomposition

Substitute the values of \(A\) and \(B\) back into the partial fraction: \[\frac{x+2}{(x+1)(x-1)} = \frac{-\frac{1}{2}}{x+1} + \frac{\frac{3}{2}}{x-1}\] This can be simplified to: \[\frac{x+2}{(x+1)(x-1)} = \frac{-1}{2(x+1)} + \frac{3}{2(x-1)}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Expression
A rational expression is simply a fraction where both the numerator and the denominator are polynomials. It extends the concept of fractions to algebraic expressions. Consider \[\frac{x+2}{(x+1)(x-1)}\]. Here, the numerator is \(x + 2\) and the denominator is the product of two polynomials, \(x+1\) and \(x-1\). This makes it a rational expression.

Rational expressions can have variables in both the numerator and the denominator, unlike regular fractions that only have numeric terms. Understanding how to manipulate these expressions is crucial in algebra, especially when simplifying them, finding common denominators, or performing operations like addition and subtraction.

To further simplify or decompose these expressions, we often use techniques like partial fraction decomposition, especially when solving calculus problems or integrating rational functions.

It's important to remember that rational expressions are undefined when the denominator equals zero. This concept leads to restrictions in the domain of these functions, often appearing in the form of vertical asymptotes in graphs.
Linear Factors
Linear factors are expressions of the form \(ax + b\), where both \(a\) and \(b\) are constants. They represent lines when graphed, hence the name 'linear'.

In the partial fraction decomposition of a rational expression, identifying linear factors is crucial because they determine the form of the decomposition. For the expression \(\frac{x+2}{(x+1)(x-1)}\), the denominator is already factored into two distinct linear factors: \((x+1)\) and \((x-1)\).

Each linear factor corresponds to a separate fraction in the decomposition, for which constants \(A\), \(B\), etc., are solved:
  • \(\frac{A}{x+1}\)
  • \(\frac{B}{x-1}\)
Understanding linear factors is not only fundamental in partial fraction decomposition but also in other areas like solving quadratic equations, graphing linear equations, and analyzing polynomial functions.

The unique characteristics of each factor, such as their roots, influence how the rational expression behaves or is simplified.
System of Equations
A system of equations consists of two or more equations with the same set of variables. Solving a system means finding values for the variables that satisfy all equations simultaneously.

In partial fraction decomposition, we often set up a system of equations to determine the unknown constants in the decomposed fractions. Let's consider the equations from the example:
  • \(A + B = 1\)
  • \(B - A = 2\)
This system arises when we compare the coefficients from both sides of the expanded equation, derived after equating and combining terms.

We solve this system using substitution or elimination methods. For these equations, adding them eliminates \(A\), allowing us to solve for \(B\). Then, substituting back to find \(A\) gives us the complete solution.

This technique not only applies to partial fraction decomposition but is also widely used in fields like physics, economics, and engineering, wherever multiple conditions are imposed on a set of variables.

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Most popular questions from this chapter

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} &y \geq(x-2)^{2}+3\\\ &y \leq-(x-1)^{2}+6 \end{aligned}$$

The break-even point for a company is the point where costs equal revenues. If both cost and revenue are expressed as linear equations, the break-even point is the solution of a linear system. In each exercise, \(C\) represents cost in dollars to produce x items, and R represents revenue in dollars from the sale of \(x\) items. Use the substitution method to find the break-even point in each case-that is, the point where \(C=R .\) Then find the value of \(C\) and \(R\) at that point. $$\begin{aligned}&C=20 x+10,000\\\&R=30 x-11,000\end{aligned}$$

Because variables appear in denominators, the system $$\begin{aligned}&\frac{5}{x}+\frac{15}{y}=16\\\&\frac{5}{x}+\frac{4}{y}=5\end{aligned}$$ is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitution-of-variable technique. Let \(t=\frac{1}{x}\) and let \(u=\frac{1}{y} \). Write a system of equations in \(t\) and \(u\) by making the appropriate substitutions.

Graph the solution set of each system of inequalities by hand. Concept Check \(\quad\) Which one of the choices that follow is a description of the solution set of the following system? $$ x^{2}+y^{2} < 36 $$ \(y < x\) A. All points outside the circle \(x^{2}+y^{2}=36\) and above the line \(y=x\) B. All points outside the circle \(x^{2}+y^{2}=36\) and below the line \(y=x\) C. All points inside the circle \(x^{2}+y^{2}=36\) and above the line \(y=x\) D. All points inside the circle \(x^{2}+y^{2}=36\) and below the line \(y=x\)

Graph the solution set of each system of inequalities by hand. $$\begin{aligned} 2 y+x & \geq-5 \\ y & \leq 3+x \\ x & \leq 0 \\ y & \leq 0 \end{aligned}$$

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