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 FactorsThe 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.

Simplification of partial fraction

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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