Let $\mathbb{F}_2 \cong \mathbb{Z}/2\mathbb{Z}$. Is $x^4+x^2+1$ irreducible in $\mathbb{F}_2[x]$? Let $\mathbb{F}_2 \cong \mathbb{Z}/2\mathbb{Z}$ denote the field of 2 elements 
(a) Is $x^4+x^2+1$ irreducible in $\mathbb{F}_2[x]$? Find a complete factorization.
(b) How many irreducible polynomials of degree 4 are there in $\mathbb{F}_2[x]$?
For the first part of the problem, I just plug in $0$ and $1$ into $x^4+x^2+1$, and know that there is no root for that polynomial, so this polynomial is irreducible. 
For the second part, I just list polynomials of degree 4 and having no zero in $\mathbb{F}_2[x]$, which are $x^4 + x+1, x^4 +x^2 +1, x^4 + x^3 +1$ , and $x^4 + x^3 +x^2 + x +1$. 
But I am not sure about whether these approaches are right, so could you help me to figure it out? Thank you!
 A: (a) no it's not: $x^4+x^2+1=(x^2+x+1)^2$
roots of $x^4+x^2+1$ are in $\mathbb{GF}_4$ which are two pair of conjugate roots so you can't see them in the $\mathbb{F}_2$.
note that $\mathbb{GF}_{2^k}$ is Galois field of order $2^k$ and also $\mathbb{F}_2=\mathbb{GF}_{2}$
(b)all irreducible polynomials of degree 4 are $x^4+x+1,x^4+x^3+1,x^4+x^3+x^2+x+1$
only way of finding is by inspection but also you can construct them using elements of $\mathbb{GF}_{16}$ by producting elements which are conjugate with respect to  $\mathbb F_{2}$. this second approach is complicated and needs $\mathbb{GF}_{16}$ elements table so it's better doing the job by inspection. 
A: The polynomial has no root in $\mathbb{F}_2$, so it has no linear factor.
The only irreducible quadratic is $x^2+x+1$. So the only way $x^4+x^2+1$ (or any quartic) can be reducible is if it is equal to $(x^2+x+1)^2$. Check whether this is the case.
To list the irreducible quartics, note that they must have shape $x^4+ax^3+bx^2+cx+1$ where the total number of $1$'s is odd. List them all, and remove $x^4+x^2+1$, since we have found it is reducible.
