Polynomial degree problem this problem is from Gelfand & Shen's Algebra book.
Problem 171. The highest coefficient of $P(x)$ is $1$, and we know that $P(1)=0,P(2)=0,P(3)=0,\ldots,P(9)=0,P(10)=0$. What is the minimal possible degree of $P(x)$? Find $P(11)$ for this case.
Answer.

 The minimal degree is $10$ and $P(11)$ is $3628800$ in this case.

I would appreciate if you could give me some hints.
 A: Hint:
When looking at a function over the real numbers $f : \mathbb{R} \to \mathbb{R}$, we have $f(k) = 0$ if and only if $(x-k) \mid f(x)$.  In other words, $f(k) = 0$ if and only if $f(x) = (x-k) g(x)$, for some function $g(x)$.  Note that in the way I worded my answer, it may or may not be true that $g(k) = 0$, but this part doesn't matter here.
A: $P(X_0) = 0 \Leftrightarrow (X - X_0) | P$
So you have $P = Q( X- 1)(...)(X-10)$ for some $Q$.
You want to minimize the degree or $P$ so you'll try to minimize the degree of $Q$ because the degree of a product is the sum of the degrees of the operands.
$-\infty$ (meaning $Q=0$) won't work because you would get $P=0$ so the highest coefficient of $P$ wouldn't be $1$.
Then you try $0$ (meaning $Q$ is a constant). The highest coefficient of a product is the product of the highest coefficients of the operands so you want $c(Q)\times1\times...\times1 = 1$ ie $c(Q)=1$ ($c(Q)$ being the highest coefficient of $Q$). Since $Q$ is constant, you get $Q=1$.
So you have $P = ( X- 1)(...)(X-10)$. And you just need to evaluate that at $11$. $P(11) = 10\times 9 \times ... \times 1 = 10! = 3,628,800$
