Multiple roots of a polynomial over a field of characteristic $p$ I have to show for what value of the prime $p$ does the polynomial $ x ^4 + x + 6$
 have a root of multiplicity $>1$ over the field of characteristic $p$.
$ p=2, 3, 5,  7 $ 
Please help. 
For $F$ a field of characteristic $3$, $f(x)= x^4 + x = x(x^3+1)$ and $f'(x) = x^3+1$. Hence, $f^′(x)= 0$ for $x=2$. Therefore in an algebraically closed field of characteristic $3$, $f(x)$ has multiple roots. 

Here's an elementary answer:
a multiple root of $x^4 + x + 6$ in $\mathbf F_p$ is the exact same thing as a root common to $x^4 + x + 6$ and its derivative $4x^3  + 1$. But, on $\mathbf Z$, we have the following formula:
$$4(x^4 + x + 6) = (4x^3 + 1) x + (3x + 24).$$
So, for $p \neq 2$, a root $\alpha$ of $x^4 + x + 6$ is multiple (in $\mathbf F_p$) iff $3\alpha + 24 = 0$.
So, for $p \neq 2, 3$, a root $\alpha$ of $x^4 + x + 6$ is multiple (in $\mathbf F_p$) iff $\alpha + 8 = 0$.
So the $p\neq 2, 3$ answering the questions are exactly the prime factors ($\neq 2, 3$) of $f(-8) = 4094 = 2\cdot 23\cdot 89$ (and, for those $p$, the multiple root is $-8$).
You have to check the two remaining cases by hand. The factorisations into prime factors of $f$ on $\mathbf F_2$ and $\mathbf F_3$ are $x(x+1)(x^2+x+1)$ and $x(x+1)^3$, respectively.
So the prime numbers answering the question are 3, 23 and 89 (and the multiple roots are $-1$ (triple), $-8$ and $-8$, respectively).
