Multiple roots in $\mathbb{Z}_p$ Let f(x) ∈ $\mathbb{Z}$[x], a polynomial of degree n.  Suppose f(x) has n distinct roots $a_1, ..., a_n$ ∈ $\mathbb{C}$.  Now, with a given f(x), we call a prime p "bad" if f(x) has a multiple factor when considered in $\mathbb{Z}_p$[x].  One example is $x^2 + x - 8$ when considered in $\mathbb{Z}_3$[x], where we have $(x+2)^2$.  Now, given f, I need to show that there can only be a finite number of bad primes.
So far I have shown that for quadratics, we only have a double root when p divides the discriminant, and thus we only have a finite number of possibilities, including zero.
I am now having trouble showing what happens when we have cubics and higher orders.
 A: Given two polynomials, if we want them to have a root in common, then their coefficients should satisfy a condition, higher the degree the more complicated this resultant polynomial (as it involves more and more coefficients.
Multiple roots for a polynomial $f$  means it is of them form $f(x) = (x-a)^2g(x)$. Working with real coefficients, and using calculus, differentiating this equation shows that the derivative $f'(x)$ has also the same number a as a root. 
That is, having a  common root with its derivative is the same as having a multiple root.  Resultant polynomial Res$\,(f,f')$ is the discriminant polynomial. 
Now coming to mod $p$ coefficients discriminant being a multiple of $p$ is the condition. Any single number can be a multiple for only a finite number of prime numbers. To know about resultant of a polynomial books written in 19th century are the best than any books on modern algebra. Books on Galois theory and field theory written in classical explicit style  could also contain this material.
