If frac{3x + 4}{x^2 - 3x + 2} = frac{A}{x - 2 | vedclass

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(B) Given the partial fraction decomposition: $\frac{3x + 4}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{B}{x - 1}$.First,factor the denominator: $x^2 - 3

Vedclass · Accepted Answer

$(10, 7)$

Vedclass · Answer

(B) Given the partial fraction decomposition: $\frac{3x + 4}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{B}{x - 1}$.First,factor the denominator: $x^2 - 3x + 2 = (x - 2)(x - 1)$.So,$\frac{3x + 4}{(x - 2)(x - 1)} = \frac{A}{x - 2} - \frac{B}{x - 1}$.Multiply both sides by $(x - 2)(x - 1)$:$3x + 4 = A(x - 1) - B(x - 2)$.To find $A$,set $x = 2$:$3(2) + 4 = A(2 - 1) - B(2 - 2) \implies 6 + 4 = A(1) \implies A = 10$.To find $B$,set $x = 1$:$3(1) + 4 = A(1 - 1) - B(1 - 2) \implies 3 + 4 = -B(-1) \implies 7 = B$.Thus,$(A, B) = (10, 7)$.

If $\frac{3x + 4}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{B}{x - 1}$,then $(A, B) = $

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