How to decompose this polynomial?! I will be really grateful if you could help me out here with decomposing this polynomial.
Decompose:
$x^{5k}+x^k+1$
This might be a child's play but well,I'm stuck a little bit.
P.S:edit the tags please if you feel not right about them.Thanks
 A: \begin{align*}x^{5k}+x^k+1&=x^{5k}-x^{2k}+(x^{2k}+x^k+1)\\
&=x^{2k}(x^k-1)(x^{2k}+x^k+1)+(x^{2k}+x^k+1)\\
&=(x^{2k}+x^k+1)(x^{3k}-x^{2k}+1)
\end{align*}
A: Do you need to factor the polynomial (that is, write it as a product of simpler polynomials), or to decompose it (that is, write it as a functional composition of simpler polynomials)?
Factoring is the more common of the operations, and ordinarily one would suspect this was what you meant (and that some kind of language barrier led you to write "decompose" instead). However, this particular polynomial seems to be tailor-made to being actually decomposed because obviously
$$ (x\mapsto x^{5k}+x^k+1) = (x\mapsto x^5+x+1)\circ(x\mapsto x^k) $$
The left composant can't be further decomposed (except trivially with linear polynomials, which are uninteresting because they are invertible). The degree of a composition of polynomials is always the product of the degrees, and 5 is prime.
Whether $x^k$ can be decomposed depends on the prime factorization of $k$. There would be one composant of $(x\mapsto x^p)$ for each prime factor of $k$, with multiplicity.
A: Set $y=x^k$.
Then, as pointed by others, $$y^5+y+1=(y^2+y+1)(y^3-y^2+1).$$
You will find the (complex) roots of the first factor easily. You will find the (real+complex pair) roots of the second factor less easily, but there are closed formulas (ask Wolfram).
Eventually, get the $x$ roots from $x=\sqrt[k]y$ (using the $k^{th}$ roots of unit) and write down the complete factorization.
