TITLE: How to factor $x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1$ by hand?
QUESTION [10 upvotes]: I know that 
$x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1 = (x^4-x^3+x+1)(x^2+2x-1)$ 
but I would not know how to do that factoring without a software. 
Some idea? Thank you!

REPLY [6 votes]: The equation is palindromic (well, almost), so:
We can write it as $$x^3\left[x^3+x^2-3x+2+\frac 3x+\frac{1}{x^2}-\frac{1}{x^3}\right]$$
$$=x^3\left[(x^3-3x+\frac 3x-\frac{1}{x^3})+\left((x-\frac 1x)^2+2\right)+2\right]$$
$$=x^3\left[u^3+u^2+4\right],$$ where $u=x-\frac 1x$
And hence the factorization is $$x^3(u+2)(u^2-u+2)$$ which will give us the expected answer.