 Indeed, because if it was $p'(x)=(a_0'+a_0'x+\dots +a_k'x^k)(b_0'+b_0'x+\dots +b_m'x^m)$ with $k,m\geq 0$, then $p(x)=(a_0+a_0x+\dots +a_kx^k)(b_0+b_0x+\dots +b_mx^m)$, so it would be in contradiction to the assumption for $p(x)$. 