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
$\begin{aligned}
& A=2, B=3, a=-4, b=4 \\
& a \cdot A+b \cdot B=-4.2+(4) \cdot 3=4
\end{aligned}$