OA is B.

OE:

We have a yes-no question, so we'll need to know if \(x\) is always prime or never prime. When working with primes, keep in mind that 2 is the smallest prime number on the only even prime. Other than that, all we know from the stem is that \(x > 0\), so let's evaluate the statements.

(1) Insufficient. Can we find any prime values for x such that \(x^3\) has 4 distinct factors? If \(x=2\), which is prime, then \(x^3 = 2^3 = 8\), which has 4 distict factors (1, ,2, 4, and 8). So it's

possible that \(x\) is prime. But does it have to be prime? Lets consider other positive numbers with 4 distinct factors. The smalelst such number is 6 (its factors are 1, 2, 3, and 6), so suppose \(x^3 = 6\). In this case, \(x = \sqrt{6}\), which is not even an integer, much less a prime number. So we've found both prime and non-prime values for \(x\) that make (1) true. Therefore, (1) is insufficent to determine whether or not \(x\) is prime. Eliminate (A) and (D).

(2)

Sufficient. In this case, we can solve the equation to find possible values for \(x\). If we subtract 6 from both sides, we get \(x^2 - x - 6 = 0\). Factor the left side using reverse FOIL to get (x-3)(x+2) = 0. So, the equation has one negative root and one positive root, meaning that there are two possible values for x, one negative and one positive. The stem tells us that \(x\) must be positive, so there is actually only one possible value of \(x\). Since we could find exactly one value for \(x\), we can answer the question definitely, if we know what \(x\) is, we know whether or not it is prime. So Statement (2) is sufficient. Answer B.

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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