8. Let $P(x)=x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$, satisfying $P(k)=k$ for $k=1,2,3,4$, then $\sqrt{P(13)-12}=$ $\qquad$ .
8. 109 Detailed Explanation: Let $Q(x)=x^{4}+a_{3} x^{3}+a_{2} x^{2}+\left(a_{1}-1\right) x+a_{0}=P(x)-x$, from $P(k)=k(k=1,2,3,4)$ we get $Q(k)=0$, and since the leading coefficient of $Q(x)$ is 1, we have $Q(x)=(x-1)(x-2)(x-3)(x-4)=\left(x^{2}-5 x+5\right)^{2}-1$, thus, $P(x)-x+1=\left(x^{2}-5 x+5\right)^{2}$. Taking $x=13$, we get $\sqrt{P(13)-12}=109$.