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$
When you multiply two polynomials, the result is the sum of each pair of terms, one from the "left" and one from the "right", multiplied together.
With $(1+x+x^2+x^3+x^4+x^5)^2$, you'll get $1*1$ once (there's only one way to pick "1" from each side). But you'll get $x$ twice, once from taking the left's $1$ and the right's $x$, and once the other way around. Hence you get a $2x$ in the product. Then there's 3 ways to get $x^2$ -- $1*x^2$, $x*x$, and $x^2*1$. It's the same from the other end, with a maximum in the middle.
Recognizing the factoring is, like many complex polynomial factorings, just a matter of being familiar with the pattern.
By direct factorization:
$$ \begin{align} P(x) &= 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} \\[5px] &= 1 + x + x^2 + x^3 + x^4 + x^5 \\ &\quad\quad + x + x^2 + x^3 + x^4 + x^5 + x^6 \\ &\quad\quad\quad\quad + x^2 + x^3 + x^4 + x^5 + x^6 + x^7\\ &\quad\quad\quad\quad\quad\quad \ldots \\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} \\[5px] &= \color{blue}{1} + x + x^2 + x^3 + x^4 + x^5 \\ &\quad\quad + \color{blue}{x}\cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\ &\quad\quad\quad\quad + \color{blue}{x^2} \cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\ &\quad\quad\quad\quad\quad\quad \ldots \\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad + \color{blue}{x^5} \cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\[5px] &= (\color{blue}{1 + x + x^2 + x^3 + x^4 + x^5}) \cdot (1 + x + x^2 + x^3 + x^4 + x^5) \end{align} $$
It is easier to expand $$(1 + x + x^2 + x^3 + x^4 + x^5)^2$$
to get $$ 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10}$$ All we have to do is check the squares and twice the products to see if the coefficients are correct.
How do we see that P(x) is a perfect square? When evaluated at different integer values of x, we get perfect squares so, that may be helpful to make an intelligent guess.