If frac{3x+1}{(x-1)(x^2+2)}=frac{A}{x-1}+frac | vedclass

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(B) Given the partial fraction decomposition: $\frac{3x+1}{(x-1)(x^2+2)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+2}$.Multiplying both sides by $(x-1)(x^2+2)

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$8C$

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(B) Given the partial fraction decomposition: $\frac{3x+1}{(x-1)(x^2+2)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+2}$.Multiplying both sides by $(x-1)(x^2+2)$,we get: $3x+1 = A(x^2+2) + (Bx+C)(x-1)$.To find $A$,set $x=1$: $3(1)+1 = A(1^2+2) \implies 4 = 3A \implies A = \frac{4}{3}$.Expanding the right side: $3x+1 = Ax^2 + 2A + Bx^2 - Bx + Cx - C = (A+B)x^2 + (C-B)x + (2A-C)$.Comparing coefficients:$x^2$ term: $A+B = 0 \implies B = -A = -\frac{4}{3}$.Constant term: $2A-C = 1 \implies 2(\frac{4}{3}) - C = 1 \implies C = \frac{8}{3} - 1 = \frac{5}{3}$.We need to find $5(A-B) = 5(\frac{4}{3} - (-\frac{4}{3})) = 5(\frac{8}{3}) = \frac{40}{3}$.Since the options provided do not match the calculated value,we check the expression $C+8 = \frac{5}{3} + 8 = \frac{29}{3}$ or $8C = \frac{40}{3}$.Thus,$5(A-B) = 8C$.

If $\frac{3x+1}{(x-1)(x^2+2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2}$,then $5(A-B)=$

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