If frac{6 x^4+13 x^3+2 x^2-x+3}{2 x^2+3 x-2}= | vedclass

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(A) Perform polynomial long division of $\frac{6 x^4+13 x^3+2 x^2-x+3}{2 x^2+3 x-2}$:$\frac{6 x^4+13 x^3+2 x^2-x+3}{2 x^2+3 x-2} = 3 x^2+2 x+1 + \frac

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(A) Perform polynomial long division of $\frac{6 x^4+13 x^3+2 x^2-x+3}{2 x^2+3 x-2}$:$\frac{6 x^4+13 x^3+2 x^2-x+3}{2 x^2+3 x-2} = 3 x^2+2 x+1 + \frac{5}{2 x^2+3 x-2}$Factor the denominator: $2 x^2+3 x-2 = (2 x-1)(x+2)$.Use partial fraction decomposition for $\frac{5}{(2 x-1)(x+2)}$:$\frac{5}{(2 x-1)(x+2)} = \frac{A}{2 x-1} + \frac{B}{x+2}$$5 = A(x+2) + B(2 x-1)$For $x = \frac{1}{2}: 5 = A(\frac{5}{2}) \implies A = 2$.For $x = -2: 5 = B(-5) \implies B = -1$.Comparing with the given form, $f(x) = 3 x^2+2 x+1$, $a = 2$, $b = 2$, $A = 2$, $B = -1$.Calculate $f(1)+a \cdot B+b \cdot A$:$f(1) = 3(1)^2+2(1)+1 = 6$.$f(1)+a \cdot B+b \cdot A = 6 + 2(-1) + 2(2) = 6 - 2 + 4 = 8$.

If $\frac{6 x^4+13 x^3+2 x^2-x+3}{2 x^2+3 x-2}=f(x)+\frac{A}{a x-1}+\frac{B}{x+b}$ then $f(1)+a \cdot B+b \cdot A=$

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