Problem 1
Question
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. $$ \frac{7}{(x-3)(x+5)} $$
Step-by-Step Solution
Verified Answer
Express as \( \frac{A}{x-3} + \frac{B}{x+5} \).
1Step 1: 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 \(x + 5\).
2Step 2: 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.
Key Concepts
Linear FactorsUnknown CoefficientsDenominator Identification
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.
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 \( x \).
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:
\[ \frac{7}{(x-3)(x+5)} = \frac{A}{x-3} + \frac{B}{x+5} \]
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.
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:\[(x-3)(x+5)\].
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.
Other exercises in this chapter
Problem 1
Compute the determinant of the given matrix. (Some of these matrices appeared in Exercises \(1-8\) in Section 8.4.) \(B=\left[\begin{array}{rr}12 & -7 \\ -5 & 3
View solution Problem 1
Find the inverse of the matrix or state that the matrix is not invertible. $$ A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] $$
View solution Problem 1
For each pair of matrices \(A\) and \(B\) in Exercises \(1-7,\) find the following, if defined \- \(3 A\) \- \(-B\) \- \(A^{2}\) \- \(A-2 B\) \- \(A B\) \- \(B
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