According to the problem, $x=1$ and $x=0$ are both double roots of $p^{\prime}(x)=0$, and $p^{\prime}(x)$ is a 4th-degree polynomial, so $p^{\prime}(x)=a x^{2}(x-1)^{2}=a\left(x^{4}-2 x^{3}+x^{2}\right) \Rightarrow p(x)=a\left(\frac{x^{5}}{5}-\frac{x^{4}}{2}+\frac{x^{3}}{3}\right)+c$. Substituting $p(0)=-1 \Rightarrow c=-1$, substituting $p(1)=1 \Rightarrow a=60$.
Therefore, $p(x)=12 x^{5}-30 x^{4}+20 x^{3}-1$.
10. Given that $p(x)$ is a 5th degree polynomial, if $x=0$ is a triple root of $p(x)+1=0$, and $x=1$ is a triple root of $p(x)-1=0$, then the expression for $p(x)$ is