Irreducible Polynomial in Field of 2 Elements? How do I show that    $ f(t) = t^2 + t +1 $ is irreducible in $K[t]$, where $K = \{0,1\}$?
I know how to tackle this over $\mathbb{Z}$ or $\mathbb{Q}$ using Guass or Eisenstein say...but I'm a little unsure how to proceed in this case.
Any help is much appreciated.
 A: Suppose $f(t)$ is reducible.(then we have to show that it is contradiction)
$f(t) = (t+a)(t+b)$ where a and b are in $K$
Case 1: $a =0,b=0$
$f(t)= t^2$. This is contradiction.
Similarly we can prove remaining cases.
Case 2: $a=1,b=1$
Case 3: $a=0,b=1$ or $a=1,b=0$
A: As said by a couple of users, this polynomial does not have a root
in $\mathbb{F}_{2}$ and since it is of degree $2$, it is irreducible.
You ask if $f$ can have a quadratic factor or a factor of higher
degree, assume that such factor $g$ exist, i.e. $f=gh$ for some
polynomial $h\neq0$. 
Then $\deg(f)=\deg(g)+\deg(h)$ implies $\deg(g)\leq \deg(f)=2$
and by our assumption $\deg(g)\geq2$ so we can not have that $f$
have a factor of higher degree. So we have $f=gh$ and since $\deg(f)=\deg(g)=2$,
it holds that $\deg(h)=0$, i.e. $h\in\mathbb{F}_{2}$ is a constant
polynomial. Since $h\neq0$ (otherwise $f=gh=0$) we have $h=1$, hence
$f=g$. 
That is you can not decompose $f$ (not to a linear factor
as you said, not to a quadratic factor or higher by this explanation) 
