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?

While others have pointed towards factorizing the expression, I would like to say that it isn't always easy for one to notice that the expression can be factorized. Also, the series has some property - the coefficients are gradually increasing or decreasing. 
Let's try to find the sum of the series because that would surely help in simplification.
$$Let, \; S = 1 + 2x + 3x^2 + 4x^3+5x^4+6x^5+5x^6+4x^7+3x^8+2x^9+x^{10} \,$$
$$ x . S = x + 2x^2 + 3x^3 + 4x^4 + 5x^5+6x^6+5x^7+4x^8+3x^9+2x^{10}+x^{11}$$
From here, we can get, 
$$S (x-1) = -(1+x+x^2+x^3+x^4+x^5) + (x^6+x^7+x^8+x^9+x^{10}) + x^{11}$$
Now use the formula of a geometric progression in the first two series to get:
$$ S (x-1) = -\frac{(x^6-1)}{x-1} + \frac{x^6(x^5 - 1)}{x-1} + x^{11}$$
$$ S = \frac{(1-x^6) + x^6(x^5 - 1) + x^{11}(x-1)}{(x-1)^2}$$
$$ S = \frac{x^{12} - 2x^6 + 1}{(x-1)^2}$$
$$ S = \frac{(x^6 - 1)^2}{(x-1)^2}$$
Now it is indeed a perfect whole square. You can simply divide $x^6-1$ by $x-1$ to get your desired result.
