How do i complete factorise x^3+6 in Q[x]?
$\displaystyle \left( x^{11} \right )^2-1=(x^{11}-1)(x^{11}+1)=p(x)\cdot q(x)$
Now by the rational roots theorem the only rational roots for each of the above binomials is $\displaystyle \pm 1$
note that $\displaystyle p(1)=(1^{11}-1)=0$ this tells us that 1 is a root of $\displaystyle x^{11}-1$ and that $\displaystyle (x-1)$ is a factor.
You can now factor this by using long division and repeat until the rational roots theorem no longer works.
I hope this helps.
p.s you will find that plus or minus one are the only rational roots.
Good luck.
Hi.
That can be done with cyclotomic polynomials:
$\displaystyle \forall k \in \mathbb{N}-\{0\}, \prod\limits_{d|k}\Phi_{d}(X)$
Furthermore, if $\displaystyle p$ is a prime number, $\displaystyle m,r \in \mathbb{N}-\{0\} $ and $\displaystyle p$ does'nt divides $\displaystyle r$, then:
$\displaystyle \Phi_{p^{m}r}(X)=\frac{\Phi_{r}(X^{p^{m}})}{\Phi_{ r}(X^{p^{m-1}})}$
As $\displaystyle \Phi_{22}(X)=\Phi_{11.2}(X)=\frac{\Phi_{2}(X^{11}) }{\Phi_{2}(X)}=\frac{X^{11}+1}{X+1}=\sum\limits_{i =0}^{i=10}(-1)^{i}X^{i}$ , and $\displaystyle \Phi_{11}(X)=\sum\limits_{i=0}^{i=10}X^{i}$ we finally have:
$\displaystyle X^{22}-1=\Phi_{1}(X)\Phi_{2}(X)\Phi_{11}(X)\Phi_{22}(X)=$$\displaystyle (X-1)(X+1)(\sum\limits_{i=0}^{i=10}X^{i})(\sum\limits _{i=0}^{i=10}(-1)^{i}X^{i})$ ,
and all those polynomials are irreducible in $\displaystyle \mathbb{Q}[X]$.