Find the first 4 elements of a given field The polynomial $x^4 + x +1$ is unsplittable under $\mathbb{Z}_2$ .
Given the following $K$: 
$K = \mathbb{Z}_2[x] / \mathbb{Z}_2[x] (x^4 + x +1)= {{a+bx+ cx^2 + dx^3 : a,b,c,d \in \mathbb{Z}_2}} $
I'm requested to find $4$ elements of sub-field of order $4$ of $K$.
The given $K$ is a field of order $16$. If $x^4 + x +1$ is unsplittable then I can't use it, I think (!??). 
If so, do I need to find some other polynomial of order 4 and try to work with it ?
Something like $t^4 - t$ maybe ? 
Regards
EDIT:
Sorry for the delay , I've some problems regarding the topic of finite fields .
First , we work with the field of $Z_{2}$ , and we know that $t^4 = t$ , and also that : 
$t^4 - t = t(t-1)(t^2 +t +1)$
As we can see , $t=0$ and $t=1$ are in the field , and now we're missing two more .
Now , assume that $t = a+bx+cx^2+dx^3$ , and place it in $t^2 +t +1$ , then :
$(a+bx+cx^2+dx^3 )^2+a+bx+cx^2+dx^3+1 = a^2+b^2 x^2+c^2 x^4+d^2 x^6+(a+bx+cx^2+dx^3)==Break=$
Since $x^4=-x-1=x+1 $  and $x^6=x^2⋅x^4=x^2(1+x)=x^2+x^3 $
$==Continues=a+bx^2+cx^4+dx^6+a+bx+cx^2+dx^3+1=a+bx^2+c(x+1)+d(x^2+x^3 )+a+bx+cx^2+dx^3+1=$
{After a lot of arithmetics} $= x(b+c+d)+x^2 (b+c)+(c+d+1)⋅1$
Finally , if b=c=1 and d=0 , we'd get that : 
$b+c+d=1+1+0=0$
$b+c=1+1=0$
$c+d+1=1+0+1=2=0$
Therefore the other two elements of this lovely field are :
$1+x+x^2  ;x^2+x$
Since they are both closed for addition and multiplication :
Addition : $ (x^2+x+1)+(x^2+x)=2x^2+2x+1=0+0+1=1$
Multiplication : $(x^2+x+1)(x^2+x)=x^4+x^3+x^3+x^2+x^2+x=x^4+2x^3+2x^2+1=x^4+0+0+x={x^4=x+1}=x+1+x=2x+1=0+1=1$
Now my questions are : 


*

*Is this correct ? 

*The addition and multiplication of the two , doesn't one of them supposed to give "0" zero ? 

*Why is it that $t^4 = t$  , always ? 
Thanks !

Further to Jyrki's comment that the multiplicative group of a finite field is cyclic: This means that you can write the $|K|-1=15$ non-zero elements as $a,a^2,\dotsc,a^{15}$, where $a$ is any of the $\phi(15)=8$ primitive elements of $K^*$. So an efficient strategy is to find the multiplicative order of some arbitrary element $a$ of $K$ other than $0$ and $1$; if it turns out to be $3$, you're done; if it turns out to be $5$, you know which $4$ elements to avoid on the second try; and if it turns out to be primitive, the multiplicative subgroup of order $3$ consists of $1=a^0,a^5$ and $a^{10}$. The easiest candidate to work with is $x$, and you can readily find that $x$, $x^2$, $x^3$, $x^4$, $x^5$ are all not $1$, so $x$ must be primitive (since otherwise its order would have to divide $15$ and thus be at most $5$). Thus the multiplicative subgroup of order $3$ consists of $1=x^0$, $x^5$ and $x^{10}$.
