Polynomial

\hspace{-16}$Here $\bf{p(x)=\frac{1}{x+1}}$ has a roots $\bf{x=0,1,2,......,11}$\\\\\\ So $\bf{(x+1).p(x)-1=0}$ has a roots $\bf{x=0,1,2,......,11}$\\\\\\ So $\bf{(x+1).p(x)-1=\mathbb{A}.x.(x-1).(x-2).(x-3)..........(x-11)...........(1)}$\\\\\\ Put $\bf{x=-1\;}$ in eqn.....$\bf{(1)\;,}$ We Get\\\\\\ $\bf{-1=\mathbb{A}.(-1).(-2)........(-12)=12!\Leftrightarrow \boxed{\mathbb{A}=-\frac{1}{12!}}}$\\\\\\ Now Put $\bf{x=12}$ in eqn........$\bf{(1)\;,}$ We Get\\\\\\ $\bf{(12+1).p(12)-1=-\frac{1}{12!}.\left\{(12).(11).(10)........(1)\right\}}$\\\\\\ So $\bf{(13).p(12)-1=-\frac{1}{12!}.12!=-1}$\\\\\\ So $\boxed{\boxed{\bf{p(12)=0}}}$