1. Answer. $\frac{x^{2}-10 x+15}{11}$.
Solution. Let the desired polynomial be $P(x)=a x^{2}+b x+c$. Then

$$
\begin{aligned}
P(x)+ & P(x+1)+\ldots+P(x+10)= \\
= & a\left(x^{2}+(x+1)^{2}+\ldots+(x+10)^{2}\right)+ \\
& \quad+b(x+(x+1)+\ldots+(x+10))+11 c= \\
= & a\left(11 x^{2}+(2+4+\ldots+20) x+\left(1+2^{2}+\ldots+10^{2}\right)\right)+ \\
& \quad+b(11 x+1+2+\ldots+10)+11 c= \\
= & 11 a x^{2}+110 a x+385 a+11 b x+55 b+11 c= \\
= & 11 a x^{2}+(110 a+11 b) x+(385 a+55 b+11 c)
\end{aligned}
$$

We obtain the equality of quadratic trinomials

$$
11 a x^{2}+(110 a+11 b) x+(385 a+55 b+11 c) \text { and } x^{2} .
$$

This is equivalent to the equality of coefficients, that is, the system of equations

$$
\left\{\begin{array}{l}
11 a=1 \\
110 a+11 b=0 \\
385 a+55 b+11 c=0
\end{array}\right.
$$

which has the unique solution $a=\frac{1}{11}, b=-\frac{10}{11}, c=\frac{15}{11}$.