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

Given $\frac{17 x-2}{12 x^2-x-20}=\frac{A}{(a x+5)}+\frac{B}{3 x+b}...(i)$ Now $\frac{17 x-2}{12 x^2-x-20}=\frac{17 x-2}{(3 x-4)(4 x+5)}$ By using partial fraction, RHS can be written as: $\begin{aligned} & \frac{17 x-2}{12 x^2-x-20}=\frac{P}{(3 x-4)}+\frac{Q}{4 x+5}...(ii) \\ & \Rightarrow 17 x-2=P(4 x+5)+Q(3 x-4)=x(4 P+3 Q)+(5 P-4 Q) \end{aligned}$ From above equation, we get: $P=2, \quad Q=3$ Putting above values in equation (ii), we get: $\frac{17 x-2}{12 x^2-x-20}=\frac{2}{(3 x-4)}+\frac{3}{4 x+5}...(iii)$ Comparing eqn (i) & (iii), we get $\begin{aligned} & A=2, B=3, a=-4, b=4 \\ & a \cdot A+b \cdot B=-4.2+(4) \cdot 3=4 \end{aligned}$