Problem Analysis #9

The function is $f(x) = 2x^3 + 10x^2 + 20x = 2x(x^2 + 5x + 10)$. For any integer $x$, $f(x)$ is a multiple of 2, so it is even. The only even prime number is 2. Therefore, for $f(x)$ to be prime, we must have $f(x) = 2$. Solving $2x^3 + 10x^2 + 20x = 2$ gives $x^3 + 5x^2 + 10x - 1 = 0$. Testing small integers shows no integer solutions (e.g., $x=0$ gives $-1$, $x=1$ gives $15$, $x=-1$ gives $-7$). Since $f(x)$ is even for all integer $x$ and never equals 2, it can never be a prime number. Thus, there are 0 such integers.

\boxed{0}