I have the field $\mathbb{F}_2(t)$, where $\mathbb{F}_2$ is the finite field, and $t$ an indeterminate. Now $X^2+t$ is an irreducible inseperable polymomial, and $X^4+t$ also.
I was looking for an irreducible inseperable polynomial of the form $a+bX^2+X^4$, where $a,b$ in $\mathbb{F}_2(t)$.
Also let's take $X^2+X+t$. How do I find the roots?
Thanks.
Let's consider $f(x)=x^2+x+t\in \mathbb{F}_2(t)[x]$. Let $E$ be a splitting field of $f(x)$ over $\mathbb{F}_2$. Let $a$ be a root of $f(x)$ in $E$. Then $f(1+a)=(1+a)^2+(1+a)+t=1+a^2+1+a+t=a^2+a+t=f(a)=0.$ Hence these are the only roots of $f(x)$ in $E$.
These roots are not elements of $\mathbb{F}_2(t)$: Indeed, if $a=\frac{b(t)}{c(t)}\in\mathbb{F}_2(t)$ with $b(t),c(t)\in\mathbb{F}_2[t]$ and $\gcd(b(t),c(t))=1$, then $a^2(t)+a(t)=-t$, therefore $b^2(t)+b(t)c(t)=-tc^2(t)$. Let $y$ be a root of $b(t)$ in its splitting field over $\mathbb{F}_2$ that's not $0$. then $b^2(y)+b(y)c(y)=-yc^2(y)$, hence $c(y)=0$. But that's a contradiction since $b(t), c(t)$ are coprime (the contradiction is that $t-y$ divides both of them). Therefore $b(t)=t^n$ for some $n\geq 1$ and $t\not| c(t)$. Suppose $n> 1$. $t^n$ divides $tc^2(t)$, therefore $t^{n-1}$ divides $c^2(t)$, hence $t$ divides $c(t)$, a contradiction. So $b(t)=t$ and our equation becomes $t^2+tc(t)=-tc^2(t)$, or equivalently $c^2(t)+c(t)=-t$, again a contradiction since this forces $c(t)$ to divide $t=b(t)$ .
For your other question, $x^4+t$ is in the form you're looking for.