If frac{17x-2}{12x^2-x-20}=frac{A}{ax+5}+frac | vedclass

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(B) Given $\frac{17x-2}{12x^2-x-20}=\frac{A}{ax+5}+\frac{B}{3x+b} \dots (i)$ Factorizing the denominator: $12x^2-x-20 = 12x^2-16x+15x-20 = 4x(3x-4)+5(

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$4$

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(B) Given $\frac{17x-2}{12x^2-x-20}=\frac{A}{ax+5}+\frac{B}{3x+b} \dots (i)$ Factorizing the denominator: $12x^2-x-20 = 12x^2-16x+15x-20 = 4x(3x-4)+5(3x-4) = (4x+5)(3x-4)$. Using partial fractions: $\frac{17x-2}{(4x+5)(3x-4)} = \frac{P}{4x+5} + \frac{Q}{3x-4}$. $17x-2 = P(3x-4) + Q(4x+5) = x(3P+4Q) + (-4P+5Q)$. Comparing coefficients: $3P+4Q = 17$ and $-4P+5Q = -2$. Solving these: $P=3, Q=2$. Thus,$\frac{17x-2}{12x^2-x-20} = \frac{3}{4x+5} + \frac{2}{3x-4}$. Comparing with $\frac{A}{ax+5} + \frac{B}{3x+b}$,we get $A=3, a=4, B=2, b=-4$. Then $a \cdot A + b \cdot B = (4)(3) + (-4)(2) = 12 - 8 = 4$.

If $\frac{17x-2}{12x^2-x-20}=\frac{A}{ax+5}+\frac{B}{3x+b}$ then $a \cdot A+b \cdot B=$

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