The question is to factorize x^4 + x to irreducible polynomials over Z2[x]. Here is my attempt but I have certain doubts:
Check roots first : p(0)= 0, so (x-0)=x is a factor;
p(1)=0, so (x-1) is a factor;
But when I check if they divide the polynomial or not, x divides it but (x-1) doesn't divide it. So, my solution in this case is x^4 +x = x(x^3 +1).
Another approach: If p(x) is reducible to g(x) and h(x), then degree(p(x))= deg(g(x)) + deg(h(x)).
So the possibilities are 4= 1+ 3 or 2+2.
But Z2 only contains polynomials of degree 0 or 1. So, I am not considering the case (2+2). If I consider the case of (1+3), then the only irreducible polynomials of degree 1 in Z2 are x and x+1 and they both divide the polynomial. So my factorization then becomes:
x^4 +x = x(x^3 +1);
x^4 +x = (x+1)(x^3-x^2+x);
But my doubt here is that both (x^3+1) and (x^3-x^2+x) are of degree 3 and they don't belong to Z2. So, my confusion is that it gives me one factorization using root method and 2 factorizations using the second approach but both of them have a 3rd degree polynomial. So, is is polynomial even reducible in Z2 or no or if I am doing something wrong here. Any suggestions or help will be appreciated.
False. There are irreducible polynomials of all degrees, even in $\mathbb Z_2$; $x^2+x+1$ is the unique degree-$2$ example. It can be shown irreducible by your second approach: if it were reducible it would have to have linear factors, which it does not.
Having found $x$ and $x-1$ as factors you have found the irreducible factors: $$x^4+x=x(x+1)(x^2+x+1)$$ Remember that $1=-1$ in $\mathbb Z_2$!