TITLE: Factorise $x^5 +x^4+1$
QUESTION [5 upvotes]: I have got the answer $(x^2+x+1)(x^3-x+1)$ , from my computer. However, even from the answer, I could not figure out what method can be used to factorise $x^5 +x^4+1$. Is there any smart or systematic method to do it?
The only way that I could think of is to consider
$x^5 +x^4+1\equiv(x^2+ax+1)(x^3+bx^2+cx+1)$ (obviously, 1 and $-1$ are not the zeros so I consider the product of a quadratic and a cubic)and then solve for $a, b, c$ . It is a feasible method but I just don't think it is a smart one.

REPLY [2 votes]: $x^5 +x^4+1\equiv(x^2+ax+1)(x^3+bx^2+cx+1)\implies$
$x^5 + x^4 + 1 = (x^5 + ax^4+ x^3)+(bx^4 + bax^3+bx^2)+(cx^3 + cax^2+cx) +(x^2 + ax + 1)=x^5 + (a+b)x^4 + (c+ba+1)x^3 + (b+ca + 1)x^2 + (c+a)x + 1\implies$
$c+a=0$ so $c=-a$
$b+ca+1=0$ so $b-a^2+1=0$ so $b=a^2-1$
$c+ba+1=0$ so $-a + a^2-a + 1=0$ so $a^2-2a+1=(a-1)^2=0$ so $a=1$ so $b=0;c=-1$
$a+b = 1$ so $1+0=1$ and
$x^5+x^4+1 = (x^2 +x + 1)(x^3-x+1)$.
Not too hard and not "not smart"
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Alternatively.   You can fill in the missing pieces.
$x^5 + x^4 + 1 = x^5 + x^4 +x^3 +x^2 + x +1 -   x^3 - x^2-x$
we can factor $x^2+x +1$ out of these componenents
$x^5 + x^4 + 1 = x^5 + x^4 +x^3 +x^2 + x +1 -   x^3 - x^2-x$
$x^3(x^2 + x + 1) + (x^2 + x+ 1) - x(x^2 +x + 1)=$
$(x^2 + x + 1)(x^3 -x +1)$