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Chapter 5: Problem 62

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Because \(x+5\) is linear and \(x^{2}-3 x+2\) is quadratic, I set up the following partial fraction decomposition: $$\frac{7 x^{2}+9 x+3}{(x+5)\left(x^{2}-3 x+2\right)}=\frac{A}{x+5}+\frac{B x+C}{x^{2}-3 x+2}$$

Short Answer

Expert verified
Yes, the statement makes sense as it is in accordance with the principle of partial fraction decomposition.

Step by step solution

01

- Understand the Concept of Partial Fraction Decomposition

Partial fraction decomposition refers to expressing a given complex fraction as a sum of simpler fractions. The given partial fraction separates \( \frac{7 x^{2}+9 x+3}{(x+5)\left(x^{2}-3 x+2\right)} \) into \(\frac{A}{x+5}\) and \(\frac{B x+C}{x^{2}-3 x+2}\). According to the rule of partial fractions, the decomposition idea is correct.
02

- Verify the Appropriateness of the Decomposition

According to the rules for decomposing partial fractions, each factor of the denominator becomes the denominator of its own fraction in the decomposed expression. In the problem, the denominator \( (x+5)\left(x^{2}-3 x+2\right) \) is separated into \( (x+5) \) and \( (x^{2}-3 x+2) \). That matches with the rule of partial fractions decomposition.
03

- Final Reasoning

The reasoning is correct. Because of the difference of their powers, the denominator can indeed be separated as per the rule of partial fractions, which is exactly what the statement shows. Hence, the statement makes sense and is accurate.

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Key Concepts

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

Linear Equation
A linear equation is an algebraic expression that represents a straight line when plotted on a graph. The standard form of a linear equation is \( ax + b = 0 \), where \( a \) and \( b \) are constants and \( x \) is the variable. Linear equations are fundamental in understanding the behavior of relationships with a constant rate of change.

When performing partial fraction decomposition, linear factors like \( x+5 \) from the denominator imply that the numerator for this part will be a constant, denoted here by \( A \). This is due to the fact that the degree of the numerator must always be less than the degree of the denominator in each fraction of the decomposition.
Quadratic Equation
A quadratic equation can be recognized by its highest degree, which is a square term. The general form is \( ax^2 + bx + c = 0 \), with \( a \), \( b \), and \( c \) being constants, and \( a \) not equal to zero. The graph of a quadratic equation is a parabola.

In the context of partial fractions, quadratic factors like \( x^{2}-3 x+2 \) in the denominator suggest that the corresponding numerator will be of a linear form, denoted as \( Bx+C \) in the exercise. The linear numerator's degree is lower than the quadratic denominator's, complying with the rules of partial fraction decomposition.
Algebraic Fractions
An algebraic fraction is a fraction in which the numerator and/or the denominator contain algebraic expressions, such as polynomials. The exercise presented involves an algebraic fraction \( \frac{7x^{2}+9x+3}{(x+5)(x^{2}-3x+2)} \).

When dealing with algebraic fractions, particularly during partial fraction decomposition, we aim to simplify the complex expression into components that are easier to integrate or differentiate if needed. This process often uncovers the underlying relationships between variables and their effects within equations.
Factoring Polynomials
The process of factoring polynomials involves breaking down a polynomial into a product of its simplest factors. This is crucial in partial fraction decomposition as we need the factors of the denominator to split the fraction appropriately.

In the exercise, the polynomial \( x^{2}-3x+2 \) appears in the denominator, which can be factored into \( (x-1)(x-2) \), exposing simpler binomial factors. Understanding how to factor polynomials is necessary to accurately decompose fractions and works hand-in-hand with recognizing different types of equations and their roles in algebraic fractions.

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Most popular questions from this chapter

When using the addition or substitution method, how can you tell if a system of linear equations has no solution? What is the relationship between the graphs of the two equations?

Sketch the graph of the solution set for the following system of inequalities: $$\left\\{\begin{array}{l} y \geq n x+b(n<0, b>0) \\ y \leq m x+b(m>0, b>0). \end{array}\right.$$

In Exercises 69鈥�70, rewrite each inequality in the system without absolute value bars. Then graph the rewritten system in rectangular coordinates. $$\left\\{\begin{array}{l} |x| \leq 1 \\ |y| \leq 2 \end{array}\right.$$

Graphing utilities can be used to shade regions in the rectangular coordinate system, thereby graphing an inequality in two variables Read the section of the user's manual for your graphing utility that describes how to shade a region. Then use your graphing utility to graph the inequalities in Exercises 97-102. $$y \leq 4 x+4$$

In \(1978,\) a ruling by the Civil Aeronautics Board allowed Federal Express to purchase larger aircraft. Federal Express's options included 20 Boeing 727 s that United Airlines was retiring and/or the French-built Dassault Fanjet Falcon \(20 .\) To aid in their decision, executives at Federal Express analyzed the following data: $$\begin{array}{lcc}\hline & \text { Boeing 727 } & \text { Falcon 20 } \\\\\hline \text { Direct Operating cost } & \$ 1400 \text { per hour } & \$ 500 \text { per hour } \\\\\text { Payload } & 42,000 \text { pounds } & 6000 \text { pounds }\end{array}$$ Federal Express was faced with the following constraints: \(\cdot\) Hourly operating cost was limited to \(\$ 35,000 .\) \(\cdot\) Total payload had to be at least \(672,000\) pounds. \(\cdot\) Only twenty 727 s were available. Given the constraints, how many of each kind of aircraft should Federal Express have purchased to maximize the number of aircraft?
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