Bernstein polynomial of $x_1^2+\cdots+x_n^2$

Question

According to S. C. Coutinho’s A Primer of Algebraic $D$-modules (see p. 95), the Bernstein polynomial of $p=x_1^2+x_2^2+\cdots+x_n^2\in K[x_1,\dots,x_n]$ (with $K$ an arbitrary field of characteristic zero) is $$B(s)=(s+1)(s+n/2).$$ Indeed, it is not hard to verify that $D=\partial_1^2+\partial_2^2+\cdots+\partial_n^2$ satisfies $Dpp^s=(s+1)(s+n/2)p^s$. Furthermore, $$D(1)=D(pp^{-1})=B(-1)p^{-1}$$ is only possible if $B(-1)=0$, showing that certainly $s+1$ divides $B(s)$. But why could we not simply have $B(s)=s+1$? Can someone find a contradiction to this (preferably, it should only rely on the theory developed in Primer up until that point)?