Is this polynomial familiar?

This polynomial with integer coefficients

$f(x)={ x }^{ 2 }+\displaystyle\prod _{ n=1 }^{ 10 }{ (x-n) }$

yields the same as given, and yet $f(11)=3628921$ . Since more than one such polynomial is possible, the answer is indeterminate.

Let $g(x) = f(x) - x^{2}$

From the problem, $g(x)$ has the root of $1,2,3,...,10$

$g(x) = (x-1)(x-2)...(x-10)h(x)$ for any polynomial $h(x)$ .

It means that $h(x)$ gives any value for any $x$ .

Therefore, $g(11)$ gives any value, and $f(11)$ also gives any value. ~~~

I feel it is impossible to determine because the question did not specify the degree of $f(x)$ .

The polynomial is not even defined before

You haven't give us the degree of the polynomial

We know that the polynomial $g(x)=f(x)-x^2$ has at least ten roots ( $x_n=n$ , where $n=1,2,\ldots,10$ ). This polynomial can yield an indefinite number of values for $f(11)$ since we don't know its degree.

Impossible to determine. We have to know the coefficient of the product of factors, and the degree of the function, in order to come to a conclusion about the value of $f(11)$ .

We have no information about the degree of $f$ . Thus the system is underdetermined. We are given ten fixed values for f. That would uniquely determine a polynomial of degree 10. But if we allow $\deg(f) \geq 11$ , then there are infinitely many solutions in the following sense. For any real number $k$ , there is a polynomial of degree 11 with the given values for $x=1..10$ and $f(11)=k$ .