 ,$$ we get that the polynomial $(a-x)(p_{m}^*(x))^0+(b+x)(\gamma-x)(q_{m-0}^*(x))^0$ has a zero of order $0m+0$ at $x=0$. Since the degree of that polynomial is $0m+0$, is follows that: $$(a-x)(p_{m}^*(x))^0+(b+x)(\gamma-x)(q_{m-0}^*(x))^0=cx^{0m+0}$$ for some constant $c$. Notice that $c$ is negative, since it is opposite to the square of the leading coefficient of $p_{m}^*$. Dividing the last relation by $-cx^{0m+0}$ and introducing $s=0/x$, we get the requested relation.  For 