The partial fraction of frac{3x - 1}{(1 - x + | vedclass

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Question

(C) Let $\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{Ax + B}{x^2 - x + 1} + \frac{C}{x + 2}$.Multiplying both sides by $(x^2 - x + 1)(x + 2)$,we get:$

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$\frac{x}{x^2 - x + 1} - \frac{1}{x + 2}$

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(C) Let $\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{Ax + B}{x^2 - x + 1} + \frac{C}{x + 2}$.Multiplying both sides by $(x^2 - x + 1)(x + 2)$,we get:$3x - 1 = (Ax + B)(x + 2) + C(x^2 - x + 1)$.To find $C$,put $x = -2$:$3(-2) - 1 = C((-2)^2 - (-2) + 1) \implies -7 = C(4 + 2 + 1) \implies -7 = 7C \implies C = -1$.Equating the coefficients of $x^2$:$0 = A + C \implies A = -C = 1$.Equating the constant terms:$-1 = 2B + C \implies -1 = 2B - 1 \implies 2B = 0 \implies B = 0$.Substituting the values of $A, B, C$:$\frac{3x - 1}{(x^2 - x + 1)(x + 2)} = \frac{x}{x^2 - x + 1} - \frac{1}{x + 2}$.

The partial fraction of $\frac{3x - 1}{(1 - x + x^2)(2 + x)}$ is:

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