Understanding 'root' in its context 
For which of the following primes p, does the polynomial $x^4+x+6$ have a root of multiplicity$> 1$ over a  field of characteristic $p$? $p=2/3/5/7$.

My book solves it using the concepts of modern algebra, which I am not very comfortable with.
I wonder if there is an intuition based method to solve this question.
Like, $x^2-2x+1$ would have $1$ as root with multiplicity$=2$. But in the given equation, everything is positive, so what is meant by root here? Is it not the value of $x$ when the graph crosses the $x$-axis?
 A: The question is slightly ambiguous, because a polynomial may only have its roots in an extension of the field where it's defined, for instance $x^2+1$ has no roots in $\mathbb{R}$, but it has roots in $\mathbb{C}$.
However, having a multiple root is equivalent to be divisible by $(x-a)^2$, where $a$ is in the field where the polynomial has its coefficients (or maybe, depending on conventions) in an extension field.
Polynomial division is carried out the same in every field:
$$
x^4+x+6 = (x-a)^2 (x^2+2ax+3a^2) + ((4a^3+1)x+(6-3a^4))
$$
where $(4a^3+1)x+(6-3a^4)$ is the remainder. For divisibility we need the remainder is zero, so
$$\begin{cases}
4a^3 + 1 = 0 \\
6 - 3a^4 = 0
\end{cases}$$
We can immediately exclude the case the characteristic is $2$, because in this case the remainder is $x+c$ ($c$ some constant term). If the characteristic is $3$, then the constant term in the remainder is zero and the first equation becomes
$$
a^3+1=0
$$
So, when $a=-1$, there is divisibility.
Note also that $0$ can never be a multiple root of the polynomial, so we can say $a\ne0$.
Assume the characteristic is neither $2$ nor $3$. We can multiply the first equation by $3a$ and the second equation by $4$; summing them up we get
$$
3a+24=0
$$
which can be simplified in $a=-8$. Plugging it in the first equation, we get
$$
4(-8)^3+1=-2047=-23\cdot89
$$
which is zero if and only if the characteristic is either $23$ or $89$.
Thus the only prime in your list that gives multiple roots is $p=3$: indeed
$$
x^4+x+6=x(x^3+1)=x(x+1)^3
$$
when the characteristic is $3$.
