Express $x^6+x^2+1$ as a product of irreducibles in $\mathbb{Z}_2[x]$

Question

I know $p(x)=x^6+x^2+1$ doesn't have any roots in $\mathbb{Z}_2[x]$ because $p(\bar{0})\neq 0$ and $p(\bar{1})\neq 0$. So it has to be either a product of a polynomial of degree $4$ and a polynomial of degree $2$ or two polynomials of degree $3$. I could try to solve $$(x^4+ax^3+bx^2+cx+d)\cdot (x^2+ex+f)=x^6+x^2+1$$ But this would take me way too long if I had to do it in a test. Is there a quicker way to do it?

Answer 1

You can use Freshman's dream to see that $$ p(x)=(x^3 + x + 1)^2. $$

Answer 2

Since all powers are even, this is the square of a polynomial $$x^6+x^2+1=(x^3+x+1)^2$$ note that $x^3+x+1$ has no roots: this means that it is irreducible. This concludes the factorization.

Answer 3

One way of getting at this if you didn't know already would be to say that if $$x^6+x^2+1=0$$ then $$x^6=x^2+1$$ and look for easy factorisations of each side. It is then easy to spot $x^2+1=(x+1)^2$ and $x^6=\left(x^3\right)^2$ and you can deploy the difference of two squares.