Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?

The coefficient of $x^n$ is the number of integer compositions of $n$ into two parts where every part has length between $0$ and $5$. You can find the coefficients with a stars and bars approach. For every $n$, draw $n$ stars in a line, and see how many places it is possible to insert one bar such that there are between $0$ and $5$ stars on either side.
The reason this works is that in the expanded form, every instance of $x^n$ results from multiplying one of $\{x^0, x^1, x^2, x^3, x^4, x^5\}$ and another one of $\{x^0, x^1, x^2, x^3, x^4, x^5\}$. Thus to find the coefficient of $x^n$, it suffices to count the number of ways to get $n$ from a sum of two members of $\{0,1,2,3,4,5\}$, where the order of the sum matters.
I do realize that my and other answers address expanding the polynomial and not factoring it. I don't think there's such an easy trick to factoring it, and I think that seeing it can be factored just comes with experience and exposure to this kind of thing. (I mean, factoring an 11-digit square number isn't really easy, so why should factoring a degree-10 square polynomial be?)
