Find $p$ values such that $p$ is prime and $f_p$ is divided by $g$ and $g \cdot x$ Let's consider the following polynomial over $\mathbb Z_p[x]$:
$$f_p = \overline{20}x^4-x^2+\overline{3}$$
I need to:


*

*find the $p$ values so that $p$ is prime and $f_p$ can be divided by
$g=x^2-x-\overline{2}$;

*find $p$ values such that $p$ is prime and $f_p$ can be divided by $h
   = x \cdot g=x^3-x^2-\overline{2}x$.


My attempt to solution
Part One.
I've divided $f_p$ by $g$ and I get a quotient $q =   \overline{20}x^2+\overline{20}x+\overline{59}$ and a remainder $r = \overline{99}x+\overline{121}$. If $f_p$ divides $g$ then its remainder must be $0$. So I need to solve:$$\overline{99}x+\overline{121} = 0 \pmod p$$ in order for that to havea solution then $d = \gcd(\overline{99}, p) \mid 121$ and that happens as $p = 11$.
Part Two
If I try to divide $f_p$ by $h$ then I get a degree 2 conguential equation, which is pretty hard to solve. So I figured that maybe I just need to divide the previous quotient by $x$, which leaves me with the easier equation:
$$20x+59 = 0 \pmod{p}$$
but the problem is $ \gcd(20,p) \mid 59 \Leftrightarrow \gcd(20,p) = 59$ and there isn't such value therefore $\nexists p$.
I have the feeling that something is quite wrong with part two of my solution, but I can't tell why, so I am asking you: am I correct? As of part one: is it also correct? Could I have got the same result in a faster way?

Method 1: do divide $f_p$ by $h$ to get the degree-2 remainder. For $h$ to divide $f_p$, every coefficient of that degree-2 remainder must be divisible by $p$ (this is a direct generalization of the previous case). So you can take the greatest common divisor of the three coefficients in the remainder.
Method 2: $h$ divides $f_p$ if and only if both $x$ and $g$ divide $f_p$ (this is because $\mathbb Z_p[x]$ is a unique factorization domain). You've already worked out the prime $p$ for which $g$ divides $f_p$. For which primes does $x$ divide $f_p$?
