CASE 3 Distinct Linear & Quadratic Factors A Bx+C ク p ax+3 X+1 12=6A la-A x2 4x+2=D2(628x13)+(Bx+C)(x+1) X4x+ 7 =2x &4x+6+BxQ+Bx+Cx +C X2_4x+4=21B)x²+ (B+C -4)x+ (6+C) Let x= -/ 123DA %3D 7ー6+C 2. ーイ+/ %3D X+1
CASE 3 Distinct Linear & Quadratic Factors A Bx+C ク p ax+3 X+1 12=6A la-A x2 4x+2=D2(628x13)+(Bx+C)(x+1) X4x+ 7 =2x &4x+6+BxQ+Bx+Cx +C X2_4x+4=21B)x²+ (B+C -4)x+ (6+C) Let x= -/ 123DA %3D 7ー6+C 2. ーイ+/ %3D X+1
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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Hi, for the part circled in green marker my teacher left out the part on how he solved for B. How do we solve for B? Thank you .
![### Case 3: Distinct Linear & Quadratic Factors
The given expression is:
\[
\frac{x^2 - 4x + 7}{(x+1)(x^2 - 2x + 3)}
\]
This is decomposed into partial fractions as follows:
\[
\frac{A}{x+1} + \frac{Bx + C}{x^2 - 2x + 3}
\]
Equating this to the original fraction:
\[
\frac{A(x^2 - 2x + 3) + (Bx + C)(x+1)}{(x+1)(x^2 - 2x + 3)}
\]
### Solving for A, B, and C:
#### Let \( x = -1 \):
Substitute \( x = -1 \) into the equation:
\[
1 = 6A \quad \Rightarrow \quad A = \frac{1}{6} \rightarrow ~\text{(correct solution in image: 2 = A)}
\]
Upon further simplification with correct substitution to match the image:
\[
x^2 - 4x + 7 = A(x^2 - 2x + 3) + (Bx + C)(x + 1)
\]
Expanding and rearranging we get:
\[
x^2 - 4x + 7 = 2x^2 - 4x + 6 + Bx^2 + Bx + Cx + C
\]
Resulting in:
\[
1 = 2 + B \quad \Rightarrow \quad B = -1
\]
Substitute into:
\[
7 = 6 + C \quad \Rightarrow \quad C = 1
\]
### Conclusion:
The partial fraction decomposition is:
\[
\frac{2}{x+1} + \frac{-x + 1}{x^2 - 2x + 3}
\]
This demonstrates the method and solution for decomposing a fraction with distinct linear and quadratic factors into its partial fractions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2Fc33379f8-5fe6-49a9-98c6-a941fc06e412%2Fg3gl41m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Case 3: Distinct Linear & Quadratic Factors
The given expression is:
\[
\frac{x^2 - 4x + 7}{(x+1)(x^2 - 2x + 3)}
\]
This is decomposed into partial fractions as follows:
\[
\frac{A}{x+1} + \frac{Bx + C}{x^2 - 2x + 3}
\]
Equating this to the original fraction:
\[
\frac{A(x^2 - 2x + 3) + (Bx + C)(x+1)}{(x+1)(x^2 - 2x + 3)}
\]
### Solving for A, B, and C:
#### Let \( x = -1 \):
Substitute \( x = -1 \) into the equation:
\[
1 = 6A \quad \Rightarrow \quad A = \frac{1}{6} \rightarrow ~\text{(correct solution in image: 2 = A)}
\]
Upon further simplification with correct substitution to match the image:
\[
x^2 - 4x + 7 = A(x^2 - 2x + 3) + (Bx + C)(x + 1)
\]
Expanding and rearranging we get:
\[
x^2 - 4x + 7 = 2x^2 - 4x + 6 + Bx^2 + Bx + Cx + C
\]
Resulting in:
\[
1 = 2 + B \quad \Rightarrow \quad B = -1
\]
Substitute into:
\[
7 = 6 + C \quad \Rightarrow \quad C = 1
\]
### Conclusion:
The partial fraction decomposition is:
\[
\frac{2}{x+1} + \frac{-x + 1}{x^2 - 2x + 3}
\]
This demonstrates the method and solution for decomposing a fraction with distinct linear and quadratic factors into its partial fractions.
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