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Chapter 8: Problem 1

In Exercises \(1-6\), find only the form needed to begin the process of partial fraction decomposition. Do not create the system of linear equations or attempt to find the actual decomposition. $$ \frac{7}{(x-3)(x+5)} $$

Short Answer

Expert verified
Express as \( \frac{A}{x-3} + \frac{B}{x+5} \).

Step by step solution

01

Identify the Denominator

Look at the given expression \( \frac{7}{(x-3)(x+5)} \). Identify that the denominator \((x-3)(x+5)\) consists of two distinct linear factors, \(x - 3\) and \(x + 5\).
02

Set Up the Partial Fractions

For distinct linear factors, set up the partial fraction decomposition using unknown coefficients. Write the expression as: \[ \frac{7}{(x-3)(x+5)} = \frac{A}{x-3} + \frac{B}{x+5} \] where \(A\) and \(B\) are constants to be determined.

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Key Concepts

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

Linear Factors
Linear factors are an important part of breaking down a complex fraction into simpler, more manageable pieces. In partial fraction decomposition, if a polynomial's denominator consists of linear factors, it simplifies the process significantly.
Linear factors are expressions of the form \( x - c \), where \( c \) is a constant. Each one corresponds to a root of the polynomial in the denominator. When you're given an expression like \( (x-3)(x+5) \), these are the linear factors.
  • The factors \( x-3 \) and \( x+5 \) imply that the roots of the polynomial are \( x = 3 \) and \( x = -5 \).
  • This indicates that the polynomial crosses or touches the x-axis at these points.
  • Linear factors are generally easy to work with because they only involve the first power of \( x \).
Identifying and expressing terms in terms of linear factors is the first big step in performing partial fraction decomposition and reveals the simplicity hidden in more complex expressions.
Unknown Coefficients
In partial fraction decomposition, after expressing the denominator in terms of linear factors, the next step involves incorporating unknown coefficients. These coefficients, represented by constants, \( A \), \( B \), etc., are crucial in rewriting the original fraction as a sum of simpler fractions.
Using the example \( \frac{7}{(x-3)(x+5)} \), we set it up as:
\[ \frac{7}{(x-3)(x+5)} = \frac{A}{x-3} + \frac{B}{x+5} \]
  • \( A \) and \( B \) are unknown coefficients that need to be determined for the equation to hold true.
  • The values of \( A \) and \( B \) ensure that when the simpler fractions are recombined, they form the original expression completely.
  • These coefficients require a system of equations to solve them, involving methods such as substitution or elimination.
It's essential to set up these unknowns accurately, as they dictate the ultimate success of the decomposition process.
Denominator Identification
Denominator identification is the foundational step in starting partial fraction decomposition. It's integral to recognize the structure of the denominator to apply the correct decomposition strategy.
When tasked with a fraction such as \( \frac{7}{(x-3)(x+5)} \), it is important to first observe the denominator:\[(x-3)(x+5)\].
  • This expression consists of two distinct linear factors: \( x-3 \) and \( x+5 \).
  • Denominator identification involves determining if these are linear, quadratic, or higher-order factors.
  • This identification guides the setup of the partial fractions that follow.
Correct denominator identification ensures that further steps in decomposition, such as creating the expression with unknown coefficients, are accurate and efficient.

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Most popular questions from this chapter

Systems of nonlinear equations show up in third semester Calculus in the midst of some really cool problems. The system below came from a problem in which we were asked to find the dimensions of a rectangular box with a volume of 1000 cubic inches that has minimal surface area. The variables \(x, y\) and \(z\) are the dimensions of the box and \(\lambda\) is called a Lagrange multiplier. With the help of your classmates, solve the system. $$ \left\\{\begin{aligned} 2 y+2 z &=\lambda y z \\ 2 x+2 z &=\lambda x z \\ 2 y+2 x &=\lambda x y \\ x y z &=1000 \end{aligned}\right. $$

Solve the given system of nonlinear equations. Use a graph to help you avoid any potential extraneous solutions. $$ \left\\{\begin{aligned} (x-2)^{2}+y^{2} &=1 \\ x^{2}+4 y^{2} &=4 \end{aligned}\right. $$

Sketch the solution to each system of nonlinear inequalities in the plane. $$ \left\\{\begin{aligned} x^{2}+y^{2} & \geq 25 \\ y-x & \leq 1 \end{aligned}\right. $$

As we stated at the beginning of this section, the technique of resolving a rational function into partial fractions is a skill needed for Calculus. However, we hope to have shown you that it is worth doing if, for no other reason, it reinforces a hefty amount of algebra. One of the common algebraic errors the authors find students make is something along the lines of $$ \frac{8}{x^{2}-9} \neq \frac{8}{x^{2}}-\frac{8}{9} $$ Think about why if the above were true, this section would have no need to exist.

In Exercises \(1-6\), find only the form needed to begin the process of partial fraction decomposition. Do not create the system of linear equations or attempt to find the actual decomposition. $$ \frac{\text { A polynomial of degree }<9}{(x+4)^{5}\left(x^{2}+1\right)^{2}} $$
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