Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$? I was at a math team meet today and one of the problems was to factor $x^5 + x^4 + x^3 + x^2 + x + 1$. It also gave the hint that it decomposes into two trinomials and a binomial.
The solution they gave was based on the fact that $\frac{x^6 - 1}{x-1} = x^5 + x^4 + x^3 + x^2 + x + 1$ and from there the solution is pretty straightforward. However, I was not aware of that factorization. The only ones I have really learned are $x^2 - y^2 = (x-y)(x+y)$ and $x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)$. Is there any other way I could have solved this factorization without using the ones they used?
 A: Here's how you could have found that identity without having known it before hand.
First, notice that setting $x = -1$ gives $0$. Thus, $(x - (-1)) = x+1$ is a root.
Using polynomial long division, we find that we can reduce it to $(x^4 + x^2 + 1)(x+1)$
Now, what to do with the $x^4 + x^2 + 1$ term? Recall that $x^3 - 1 = (x-1)(x^2 + x + 1)$. If we plug $x = x^2$ into this identity, we find $x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1)$, so $x^4 + x^2 + 1 = \frac{x^6 - 1}{x^2 - 1}$.
We have now reduced it to $\frac{(x^6 - 1)(x+1)}{(x-1)(x+1)}$ = $\frac{x^6 - 1}{x-1}$ and, as you said, the rest is easy from here using the identities you listed in the OP.
A: You should be(come) aware of:  $\frac{x^n-1}{x-1} = \sum_{k=0}^{n-1} x^k$
It's quite useful.
$$x^5+x^4+x^3+x^2+x+1 \\ = \frac{x^6-1}{x-1} \\ = \frac{(x^3-1)(x^3+1)}{x-1} \\ = (x^2+x+1)(x^3+1) \\ = (x^2+x+1)(x+1)(x^2-x+1)$$
The only other way would be to guess -1 as a root, because you have six terms in ascending polynomial sequence. 
$$ x^5+x^4+x^3+x^2+x+1 = (x+1)(x^4+x^2+1)$$
Then split the resulting tetranomial, by solving: $$\exists a, b : x^4 + x^2 + 1 = (x^2+ax+1)(x^2+bx+1) \\ (x^2+a x + 1)(x^2+b x +1) = x^4 + (a+b) x^3 + (2+ab) x^2 + (a+b)x + 1 \\ \therefore a+b=0 \land ab=-1 \\ \therefore a = \pm 1, b=\mp 1 $$
And so:
$$ x^5+x^4+x^3+x^2+x+1 = (x+1)(x^2+x+1)(x^2-x+1)$$
A: You can still factor this. Notice that 
\begin{align*}
x^5+x^4+x^3+x^2+x+1 &= x^3(x^2+x+1)+1(x^2+x+1) \\
&=(x^3+1)(x^2+x+1) \\
&=(x+1)(x^2-x+1)(x^2+x+1).
\end{align*}
Now you can go one step further and show that both $(x^2-x+1)$ and $(x^2+x+1)$ are irreducible in the real numbers just by using the discriminant. 
