Problem Analysis #9

To determine the number of integers $x$ for which $f(x) = 2x^3 + 10x^2 + 20x$ is prime, we start by factoring the polynomial:

For $f(x)$ to be prime, one of the factors must be $\pm 1$ and the other must be a prime number. We analyze each factor separately:

1. Factor $2x$:

   • If $2x = 1$, then $x = \frac{1}{2}$, which is not an integer.
   • If $2x = -1$, then $x = -\frac{1}{2}$, which is also not an integer.
   • If $2x = 2$, then $x = 1$.
   • If $2x = -2$, then $x = -1$.

2. Factor $x^2 + 5x + 10$:

   • This quadratic expression is always positive for all real $x$ because its discriminant $25 - 40 = -15$ is negative, indicating no real roots. Thus, $x^2 + 5x + 10 > 0$ for all $x$.

Next, we check the values of $x$ that make $2x$ a prime number:

• $x = 1$: $2x = 2$, which is prime.
• $x = -1$: $2x = -2$, which is not prime.

For $x = 1$:

which is not prime.

For $x = -1$:

which is not prime.

We also check other integer values of $x$ to ensure no other primes are generated:

• For $x = 0$, $f(0) = 0$, which is not prime.
• For $x = 2$, $f(2) = 2(8) + 10(4) + 20(2) = 16 + 40 + 40 = 96$, which is not prime.
• For $x = -2$, $f(-2) = 2(-8) + 10(4) + 20(-2) = -16 + 40 - 40 = -16$, which is not prime.

After checking all possible integer values, we find that none of them result in a prime number for $f(x)$.

Thus, the number of integers $x$ for which $f(x)$ is prime is $\boxed{0}$.