#2 Form of partial fractions. 3x+1 A G-2)(x243) X-2 Bx +C x² +3 Find Constants A, B, and C

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Topic: Partial Fraction Decomposition

This set of notes demonstrates the process of partial fraction decomposition with a specific example. The goal is to find the constants \(A\), \(B\), and \(C\).

#### Given Equation:
\[
\frac{3x + 1}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3}
\]

#### Steps:

1. **Express the Equation:**
   \[
   3x + 1 = A(x + 3) + B(x - 2)
   \]

2. **Expand and Combine Like Terms:**
   \[
   3x + 1 = (A + B)x + (3A - 2B)
   \]

3. **Setup the System of Equations:**
   - By comparing coefficients, we have:
     - \( A + B = 3 \)
     - \( 3A - 2B = 1 \)

4. **Solve for Constants:**

   - **Using \( x = 0 \):**
     Plugging \( x = 0 \) into the equation:
     \[
     1 = 3A + 2C \implies 1 = 3A + 2C = 2
     \]
     - Simplifying gives \( C \).

   - **Using \( x = 2 \):**
     Plugging \( x = 2 \) into the equation:
     \[
     7 = A(7) \implies A = 1
     \]

   - **Using \( x = 1 \):**
     Plugging \( x = 1 \) into the equation:
     \[
     4 = 5 - B \implies B = 1
     \]

#### Final Solution:
The constants are determined as \( A = 1 \), \( B = 1 \), and \( C \) is found from the equation system.

These values are boxed and highlighted in the solution as follows:
- \( A = 1 \)
- \( B = 1 \)
- Appropriate calculations are shown, with equations and simplifications.

This example demonstrates how to decompose a rational function into partial fractions by solving a system of equations derived from equating coefficients.
Transcribed Image Text:### Topic: Partial Fraction Decomposition This set of notes demonstrates the process of partial fraction decomposition with a specific example. The goal is to find the constants \(A\), \(B\), and \(C\). #### Given Equation: \[ \frac{3x + 1}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3} \] #### Steps: 1. **Express the Equation:** \[ 3x + 1 = A(x + 3) + B(x - 2) \] 2. **Expand and Combine Like Terms:** \[ 3x + 1 = (A + B)x + (3A - 2B) \] 3. **Setup the System of Equations:** - By comparing coefficients, we have: - \( A + B = 3 \) - \( 3A - 2B = 1 \) 4. **Solve for Constants:** - **Using \( x = 0 \):** Plugging \( x = 0 \) into the equation: \[ 1 = 3A + 2C \implies 1 = 3A + 2C = 2 \] - Simplifying gives \( C \). - **Using \( x = 2 \):** Plugging \( x = 2 \) into the equation: \[ 7 = A(7) \implies A = 1 \] - **Using \( x = 1 \):** Plugging \( x = 1 \) into the equation: \[ 4 = 5 - B \implies B = 1 \] #### Final Solution: The constants are determined as \( A = 1 \), \( B = 1 \), and \( C \) is found from the equation system. These values are boxed and highlighted in the solution as follows: - \( A = 1 \) - \( B = 1 \) - Appropriate calculations are shown, with equations and simplifications. This example demonstrates how to decompose a rational function into partial fractions by solving a system of equations derived from equating coefficients.
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