$\begin{array}{l}(x-2)(x-2)(x+1)-3(x-2)(x+1)=0 \\ (x-2)(x+1)[(x-2)-3]=0 \\ (x-2)(x+1)(x-5)=0 \\ x=2,-1 \text { or } 5 \\ \Rightarrow \text { Number of positive roots }=c=2\end{array}$
I1.3 If $c$ is the total number of positive roots of the equation $(x-b)(x-2)(x+1)=3(x-b)(x+1)$, find the value of $c$.