$(b+x)(p_{m-0}^*(x))^0+(a-x)(\gamma-x)(q_{m-0}^*(x))^0$ has a zero of order $0m$ at $x=0$. Since the degree of that polynomial is $0m$, is follows that: $$(b+x)(p_{m-0}^*(x))^0+(a-x)(\gamma-x)(q_{m-0}^*(x))^0=cx^{0m},$$ for some constant $c$. Notice that $c$ is positive, since it equals t