Problem Analysis #9

To determine how many integers $x$ make the function

$$f(x) = 2x^3 + 10x^2 + 20x$$

a prime number, we begin by analyzing the structure of the function.

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Step 1: Factor the Expression

We can factor out $2x$ from all terms:

$$f(x) = 2x(x^2 + 5x + 10)$$

This is key because it shows that $f(x)$ is always even for any integer $x$. Why?

• The term $2x$ is even if $x$ is an integer.
• The quadratic $x^2 + 5x + 10$ is always an integer for integer $x$.
• Therefore, the product $2x(x^2 + 5x + 10)$ is always even.

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Step 2: Analyze the Nature of Primes

Recall that a prime number must be:

• A positive integer
• Greater than 1
• Have exactly two positive divisors: 1 and itself

Moreover, the only even prime number is 2. All other even numbers are divisible by 2 and thus composite.

So, for $f(x)$ to be prime, it must equal exactly 2, since $f(x)$ is always even.

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Step 3: Solve for When $f(x) = 2$

Set up the equation:

$$2x^3 + 10x^2 + 20x = 2$$

Divide both sides by 2:

$$x^3 + 5x^2 + 10x = 1$$

Now check for integer solutions to this equation. Try small integer values of $x$:

| $x$ | $x^3 + 5x^2 + 10x$ | |--------|-------------------------| | -5 | -125 + 125 - 50 = -50 | | -4 | -64 + 80 - 40 = -24 | | -3 | -27 + 45 - 30 = -12 | | -2 | -8 + 20 - 20 = -8 | | -1 | -1 + 5 - 10 = -6 | | 0 | 0 | | 1 | 1 + 5 + 10 = 16 | | 2 | 8 + 20 + 20 = 48 |

None of these values equal 1. Thus, there is no integer $x$ such that $f(x) = 2$.

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Step 4: Consider Other Possibilities

Since $f(x)$ is always even, and the only even prime is 2, which is never achieved with integer $x$, it follows that:

• For all integer $x$, $f(x)$ is either:
  • Negative (when $x < 0$), or
  • Zero (when $x = 0$), or
  • Positive even numbers greater than 2 (when $x > 0$)

In all cases, $f(x)$ is not a prime number.

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Final Conclusion

There are no integer values of $x$ for which $f(x)$ is a prime number.

$$\boxed{0}$$