<s>[INST] 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?
 [/INST] By direct factorization:
$$
\begin{align}
P(x) &= 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} \\[5px]
 &= 1 + x + x^2 + x^3 + x^4 + x^5 \\
 &\quad\quad + x + x^2 + x^3 + x^4 + x^5 + x^6 \\
 &\quad\quad\quad\quad + x^2 + x^3 + x^4 + x^5 + x^6 + x^7\\
 &\quad\quad\quad\quad\quad\quad \ldots \\
 &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad  + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10}  \\[5px]
 &= \color{blue}{1} .(1+ x + x^2 + x^3 + x^4 + x^5) \\
 &\quad\quad + \color{blue}{x}\cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\
 &\quad\quad\quad\quad + \color{blue}{x^2} \cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\
 &\quad\quad\quad\quad\quad\quad \ldots \\
 &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad  + \color{blue}{x^5} \cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\[5px]
 &= (\color{blue}{1 + x + x^2 + x^3 + x^4 + x^5}) \cdot (1 + x + x^2 + x^3 + x^4 + x^5)
\end{align}
$$
</s>