Problem Analysis #9

To find the number of integers $x$ for which $f(x) = 2x^3 + 10x^2 + 20x$ is prime, we can factor the expression:

Since $x$ is an integer, $x(x^2 + 5x + 10)$ is also an integer. This means $f(x)$ is always an even integer.

The only even prime number is $2$. Therefore, for $f(x)$ to be prime, we must have:

Dividing both sides by $2$ gives:

Since $x$ must be an integer, the factors $x$ and $x^2 + 5x + 10$ must be integers that multiply to $1$. The only integer divisors of $1$ are $1$ and $-1$. This gives us two cases to check:

1. Case 1: $x = 1$ Substituting $x = 1$ into the second factor:

   $$1^2 + 5(1) + 10 = 16 \neq 1$$

   So, $x = 1$ is not a solution.

2. Case 2: $x = -1$ Substituting $x = -1$ into the second factor:

   $$(-1)^2 + 5(-1) + 10 = 1 - 5 + 10 = 6 \neq -1$$

   So, $x = -1$ is not a solution.

Since there are no integer solutions to the equation, there are no integers $x$ for which $f(x)$ is prime.

\boxed{0}