Let $f(x)=Q(x)R(x)$. Then taking mod 11, $f(x)=x^m(x-10)^n (x+10)^n$. Since $\mathbb{Z}_{11}[x]$ is a UFD, it follows that $Q(x)\equiv x^a (x-10)^b (x+10)^c (\bmod\; 11)$, and similar conclusion for $R(x)$. Plugging in $x=0$ suggests not both of $Q(0),R(0)$ are divisible by 11, so it follows that $a=0$ or $a=m$. Plugging in $x=10$ suggests $b=0$ or $b=m$. Same thing for $c$. However, $Q(10)\equiv Q(-10) (\bmod\; 20)$ and the only possibilities for these numbers are $1,-1,-11,11$ so it follows that $Q(10)=Q(-10)$. If $b=m$ then it follows that $11\mid Q(10)$ so $11\mid Q(-10)$ so $c=m$ as well. Therefore, either $Q(x)=x^m(x^2-100^n)+11g(x)$ or $Q(x)=x^m+11g(x), R(x)=(x^2-100)^n + 11h(x)$. The first case is impossible because $\deg(P)\ge \deg (f)$. It remains to analyze the second case. We have $11g(x)(x^2-100)^n + 11h(x)x^m + 121g(x)h(x)=-11$ $g(x)(x^2-100)^n + h(x)x^m + 11g(x)h(x)=-1$ Furthermore, we know $|Q(10)|\le 11$ so $|h(10)|=|h(-10)|=1$. Similarly, $|g(0)|=1$ Plug $x=-10$ gives $h(-10) \equiv 1(\bmod\; 11)$, so $h(-10)=-1$. We have $-1 (-10)^m + 11g(-10)h(-10)=-1$, so $g(-10)$ is 1 mod 10. Furthermore, $m$ is even and $g(0)=1$. Observe $Q(x)$ and $R(x)$ are irreducible by noting that $Q,R$ is the only way to split $f$. Since $m$ is even, it follows that $Q(-x)R(-x)=Q(x)R(x)$ so it follows that $\{ Q(-x), R(-x) \} = \{Q(x), R(x)\}$. Since $\deg Q$ is even and $\deg R$ is odd, $Q(-x)=Q(x)=S(x^2)$. Thus, letting $m=2k$, we have $x^{2k}(x^2-100)^n-11=S(x^2)T(x^2)$ Hence $x^k(x-100)^n-11=S(x)T(x)$. We mod 11 again! Redefine $g,h$ st $S(x)=x^k+11g(x), T(x)=(x-100)^n+11h(x)$. Since $S(x^2)=Q(x)$, it follows that $S(0)=11$ and $T(0)=-1$. Furthermore, we have $S(100)=Q(10)=1$. However, $100\mid S(100)-S(0)=10$, contradiction!!!!!!!!!