About Spliting Field What is splitting field of polynomial $X^3+X+\bar 1$ in $\mathbb{F_5}$.
Attempt: To show this we first want to check irreducibility which is clear in case of finite field since there are only 5 elements so put values and check it has no root in $\mathbb{F_5}$. After that how to proceed?

My answer to this question with software is in the following form.
First, I find a positive integer $n$ such that in the decomposition of $x^n-1$ in modulo $5$  we have the factor $x^3+x+1$. In your question we get $n=62$. In fact, we get 
$$
(x^{62}-1) \pmod{5}=
\left( {x}^{3}+2\,x+4 \right)  \left( {x}^{3}+2\,x+1 \right)  \left( 
{x}^{3}+{x}^{2}+3\,x+4 \right)  \left( {x}^{3}+4\,{x}^{2}+x+1 \right) 
 \left( x+1 \right)  \left( {x}^{3}+3\,{x}^{2}+4 \right)  \left( {x}^{
3}+{x}^{2}+3\,x+1 \right)  \left( {x}^{3}+2\,{x}^{2}+x+4 \right) 
 \left( {x}^{3}+{x}^{2}+1 \right)  \left( {x}^{3}+3\,{x}^{2}+4\,x+1
 \right)  \left( {x}^{3}+x+4 \right)  \left( {x}^{3}+4\,{x}^{2}+3\,x+4
 \right)  \left( {x}^{3}+3\,{x}^{2}+x+1 \right)  \left( {x}^{3}+x+1
 \right)  \left( {x}^{3}+{x}^{2}+4\,x+1 \right)  \left( {x}^{3}+4\,{x}
^{2}+4 \right)  \left( {x}^{3}+{x}^{2}+x+4 \right)  \left( x+4
 \right)  \left( {x}^{3}+4\,{x}^{2}+3\,x+1 \right)  \left( {x}^{3}+2\,
{x}^{2}+1 \right)  \left( {x}^{3}+2\,{x}^{2}+4\,x+4 \right)  \left( {x
}^{3}+4\,{x}^{2}+4\,x+4 \right) 
$$
Then, we obtain a positive integer $k$ such that $62\mid 5^k-1$ which is $k=3$. Then, we construct $\mathbb{F}_{5^3}$ with any irreducible polynomial such as $\bf f$.
Finally, we choose  elements of the field $\mathbb{F}_{5^3}$ that their order is $62$ and test which of these elements satisfy $x^3+x+1=0$. 
