The question is this: What is the number and degrees of the irreducible factors of $f(x)=x^{11}+1$ in $\mathbb{F}_{2^n}[x]$, for $n\in \mathbb{N}$?
Thoughts:
I know that the elements of $\mathbb{F}_{2^n}$ are precisely roots of the polynomial $x^{2^n}-x$ over $\mathbb{F}_2.$ Next, if $n=1$ then obviously $f(1)=0$, so that $f(x)=(x+1)g(x)$ where $\deg g(x)=10$.
One approach would be to figure out the degree of the splitting field of $f(x)$ over $\mathbb{F}_{2^n}$, but I'm not really sure how to do this since $2^n$ is not a prime if $n>1$. (I can do it if $n=1$, which i also think is the only case where $f(x)$ is reducible).
Next, I can show that $f(x)$ cannot have any roots in $\mathbb{F}_{2^n}$, for $n>1$.
Suppose that $f(a)=0;$ that is, $a^{11}\equiv -1\mod 2^n$, and so also $a^{22}\equiv 1\mod 2.$
Since $a\in \mathbb{F}_{2^n}$ it follows that $a^{2^n-1}\equiv 1\mod 2$, as well.
So, it follows that $22|2^n-1$. That is, there is some $m$ such that $2^n-1=2(11m)$, which is not possible for $n>1$ since $2^n-1$ is odd.
So, this only tells us that $f(x)$ does not have a linear factor in $\mathbb{F}_{2^n}[x]$, but not sure how to proceed from here?
Thanks!
Because we are in characteristic two, we have the factorization $$ f(x)=x^{11}+1=(x+1)(x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1) $$ over the prime field $\mathbb{F}_2$. The question really is about the factorization of that degree ten cyclotomic polynomial $\phi_{11}(x)$. Its roots are eleventh primitive roots of unity. If $\beta$ is one of them (in some extension field $\mathbb{F}_{2^t}$, then the others are $\beta^j,j=2,3,\ldots,10.$
The first thing we see is that $\phi_{11}(x)$ is irreducible in $\mathbb{F}_2[x]$. This is because if $m(x)$ is the minimal polynomial of $\beta$, then for any root $\gamma$ of $m(x)$ we also have that its Frobenius conjugate $\gamma^2$ is also a root. (If you haven't seen the Frobenius automorphism yet, then ask.) So $m(x)$ necessarily has, in addition $\beta$ also $\beta^2$, $\beta^4$, $\beta^8$, $\beta^{16}=\beta^{11+5}=\beta^5$ et cetera as roots. Because $2$ generates the multiplicative group $\mathbb{Z}_{11}^*$ (check this if you haven't seen it!), we see that all the primitive eleventh roots of unity must be zeros of $m(x)$, and thus $m(x)=\phi_{11}(x)$. The claim follows.
So all the roots $\beta^j, j=1,2,\ldots,10,$ are in an extension of degree ten, i.e. in the field $\mathbb{F}_{1024}$. Therefore the splitting field of $\phi_{11}(x)$ over $\mathbb{F}_{2^n}$ is the compositum of $\mathbb{F}_{2^n}$ and $\mathbb{F}_{1024}$, or the field $\mathbb{F}_{2^m}$ where $m$ is the least common multiple of $n$ and $10$. After all, this is the smallest field that contains the field $\mathbb{F}_{1024}$ as well as the field $\mathbb{F}_{2^n}$. A consequence of this is that $\phi_{11}(x)$ remains irreducible over $\mathbb{F}_{2^n}$ whenever $\gcd(n,10)=1$.
A few words about the factors of $\phi_{11}(x)$ in the cases $\gcd(n,10)>1$. When we study the minimal polynomial $m_{j,n}(x)$ of $\beta^j$ over $\mathbb{F}_{2^n}$ the above rule changes to the form: if $\gamma$ is a zero of $m_{j,n}(x)$, then the element $\gamma^{2^n}$ is also a zero of $m_{j,n}(x)$. For example over $\mathbb{F}_4$ we see that the zeros of $m_{1,2}(x)$ are $$ \beta, \beta^4, \beta^{16}=\beta^5, \beta^{5\cdot4}=\beta^9, \beta^{9\cdot4}=\beta^3, $$ but we stop here, because $\beta^{3\cdot4}=\beta$. So $$ m_{1,2}(x)=(x-\beta)(x-\beta^4)(x-\beta^5)(x-\beta^9)(x-\beta^3) $$ has coefficients in $\mathbb{F}_{4}$ and is thus a quintic factor of $\phi_{11}(x)$ over $\mathbb{F}_{4}$. The other factor $m_{-1,2}(x)$ is the reciprocal polynomial of $m_{1,2}(x)$, i.e. its zeros are the inverses of those of $m_{1,2}(x)$. In light of the above, the factorization of $\phi_{11}(x)$ in to irreducibles over $\mathbb{F}_{2^n}$ is $\phi_{11}=m_{1,2} m_{-1,2}$ whenever $\gcd(n,10)=2$.
Of course, when $10\mid n$, the polynomial $\phi_{11}(x)$ splits into linear factors over $\mathbb{F}_{2^n}$.
The remaining case is that of $\gcd(n,10)=5$. Because $(\beta^j)^{32}=\beta^{-j}$, the polynomial $$ m_{j,5}(x)=(x-\beta^j)(x-\beta^{-j})=x^2-x(\beta^j+\beta^{-j})+1 $$ has coefficients in $\mathbb{F}_{32}$ for all $j=1,2,3,4,5$. This means that over $\mathbb{F}_{32}$ the polynomial $\phi_{11}(x)$ splits into a product of five irreducible quadratic factors. The same factorization obviously is there, whenever $\gcd(n,10)=5$.