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?
 A: By direct factorization:
$$
\begin{align}
P(x) &= 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} \\[5px]
 &= 1 + x + x^2 + x^3 + x^4 + x^5 \\
 &\quad\quad + x + x^2 + x^3 + x^4 + x^5 + x^6 \\
 &\quad\quad\quad\quad + x^2 + x^3 + x^4 + x^5 + x^6 + x^7\\
 &\quad\quad\quad\quad\quad\quad \ldots \\
 &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad  + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10}  \\[5px]
 &= \color{blue}{1} .(1+ x + x^2 + x^3 + x^4 + x^5) \\
 &\quad\quad + \color{blue}{x}\cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\
 &\quad\quad\quad\quad + \color{blue}{x^2} \cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\
 &\quad\quad\quad\quad\quad\quad \ldots \\
 &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad  + \color{blue}{x^5} \cdot(1 + x + x^2 + x^3 + x^4 + x^5) \\[5px]
 &= (\color{blue}{1 + x + x^2 + x^3 + x^4 + x^5}) \cdot (1 + x + x^2 + x^3 + x^4 + x^5)
\end{align}
$$
A: 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.
A: 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.   
A: Use the distributive property.
$$(1 + x + x^2 + x^3 + x^4 + x^5)(1 + x + x^2 + x^3 + x^4 + x^5) $$
is equal to
$$
    \begin{matrix}
    (1)(1 + x + x^2 + x^3 + x^4 + x^5) \\
     (x)(1 + x + x^2 + x^3 + x^4 + x^5) \\
     (x^2)(1 + x + x^2 + x^3 + x^4 + x^5) \\
     (x^3)(1 + x + x^2 + x^3 + x^4 + x^5) \\
     (x^4)(1 + x + x^2 + x^3 + x^4 + x^5) \\
     (x^5)(1 + x + x^2 + x^3 + x^4 + x^5) \\ 
    \end{matrix}
$$
When the above is expanded, we get $36$ products. Arrange the results of these $36$ products into columns whose entries have the same degree.
$$
    \begin{matrix}
    1 & x & x^2 & x^3 & x^4 & x^5 \\
     & x & x^2 & x^3 & x^4 & x^5 & x^6 \\
     &  & x^2 & x^3 & x^4 & x^5 & x^6 & x^7 \\
     &  & & x^3 & x^4 & x^5 & x^6 & x^7 & x^8 \\
     &  & & & x^4 & x^5 & x^6 & x^7 & x^8 & x^9 \\
     &  & & & & x^5 & x^6 & x^7 & x^8 & x^9 & x^{10} \\ 
    \end{matrix}
$$
Adding like powers of $x$ shows how the pattern of coefficients arises, giving
$$= \;\;\; 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + 5x^6 + 4x^7 + 3x^8 + 2x^9 + x^{10} $$
In the case of $\;(1 + x + x^2 + \ldots + x^{n-1} + x^n)^2,\;$ the same pattern of coefficients --- increasing by increments of $1$ from $1$ to some maximum value, and then decreasing by increments of $1$ from the maximum value to $1$ --- can be seen by thinking about the following.
$$
    \begin{matrix}
    1 & x & x^2 & x^3 & x^4 & \ldots & x^{n-1} & x^n \\
     & x & x^2 & x^3 & x^4 & \ldots & x^{n-1} & x^n & x^{n+1} \\
     &  & x^2 & x^3 & x^4 & \ldots & x^{n-1} & x^n & x^{n+1} & x^{n+2} \\
     &  & & x^3 & x^4 & \ldots & x^{n-1} & x^n & x^{n+1} & x^{n+2} & x^{n+3} \\
     &  & & & x^4 & \ldots & x^{n-1} & x^n & x^{n+1} & x^{n+2} & x^{n+3} & x^{n+4} \\
    \end{matrix}
$$
$$\cdot$$
$$\cdot$$
$$\cdot$$
A: 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.
A: The coefficient of $x^n$ ($0\leq n\leq 10$) in the expansion is the number of solutions to
$$
x_{1}+x_{2}=n;\quad 0\leq x_i\leq5.
$$
We can compute this via inclusion exclusion. Let $U$ be the set of all solutions to the previous question in non-negative integers and $A_{i}$ be the set of solutions in $U$ which $x_i>5$. Then
$$
\begin{align}
|A_{1}^c\cap A_{2}^c|&=|U|-|A_{1}|-|A_{2}|+|A_{1}\cap A_{2}|\\
&=(n+1)-2(n-5)[n\geq6]+0
\end{align}
$$
where $[\cdot]$ is the iverson bracket since $A_{1}\cap A_{2}=\varnothing$.
A: This is equivalent to multiplying $111111 \times 111111$.  There is a principle in logic called "universal generalization".  Since no property of the $10$ in $10^k$ the base of $11111$ is being used, because there are no carries, it can be generalized to $x^k$.
A: 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?)
