How can I factorize $x^{10}+x^5+1$?

Question

How can I factorize $x^{10}+x^5+1$ ?

hope you explain the steps :)

Thanks

Answer 1

In case you're factoring over $\mathbb{Q}$:

Write $x^{15} - 1 = (x^{10} + x^5 + 1)(x^5 - 1)$ and note that $x^3 - 1$ divides $x^{15} - 1$ but not $x^5 - 1$. It turns out that the factor $x^2 + x + 1$ of $x^3 - 1$ divides $x^{10} + x^5 + 1$. Long division then gives $x^{10} + x^5 + 1 = (x^2 + x + 1)(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$. The right hand side has degree $8 = \phi(15)$, so it's the $15$th cyclotomic polynomial. So it's irreducible and we are done.

Answer 2

Hint:

$x^{10} + x^5 + 1 = (x^5)^2 + x^5 + 1 = y^2 + y + 1 = (y - root_1)(y - root_2)$

if $y = x^5$

Answer 3

Hint: Setting $t=x^{5}$ you get that $$ x^{10}+x^{5}+1=t^{2}+t+1 $$

Answer 4

the factorization $$ x^{15} - 1 = (x^5 -1)(x^{10}+x^5+1) $$ shows that the factors you want are $x-\zeta$ where $\zeta$ ranges over the set of $10$ fifteenth-roots of unity which are not $5$th roots of unity. so the factorization may be expressed as: $$ x^{10}+x^5 + 1 = \prod_{(k,j) \in \{1,2\}\times\{0,1,2,3,4\}} \left(x-e^{(5k+3j)\frac{2\pi i}{15}} \right) $$

Answer 5

You could observe that $x^{2}+x + 1 = (x- \alpha)(x- \alpha^{2})$ where $\alpha = \exp(\frac{2 \pi i}{3})$, so that $x^{10}+x^{5} + 1 = (x^{5}- \alpha)(x^{5}- \alpha^{2})$. Then note that that $x^{5} - \alpha = (x - \beta \alpha^{2})(x - \beta^{2} \alpha^{2})(x - \beta^{3} \alpha^{2})(x - \beta^{4} \alpha^{2}),$ where $\beta = \exp(\frac{2 \pi i}{5})$, and $x^{5} - \alpha^{2} = (x - \beta \alpha)(x - \beta^{2} \alpha)(x - \beta^{3} \alpha)(x - \beta^{4} \alpha),$ leading to $ (x^{10}+x^{5}+1) = (x - \beta \alpha^{2})(x - \beta^{2} \alpha^{2})(x - \beta^{3} \alpha^{2})(x - \beta^{4} \alpha^{2})$ $(x - \beta \alpha)(x - \beta^{2} \alpha)(x - \beta^{3} \alpha)(x - \beta^{4} \alpha)$.

This is basically the same as David Holden's answer, but derived via a slightly different route.