Hi guys, could someone explain why the second option isn’t an answer?

If we add 3 to (x-1), it becomes (x+2); if we subtract 3 to (x+1), it becomes (x-2); so it will be a multiple of 3. By extension, why can’t (x-5) be an option if we just subtract 3 from (x-2)?

The final product does have the factor (x-5), but it does not have the factor (x+4) as shown in the second answer choice. In fact, it doesn’t seem that we can make this factor (x+4) by +3 or -3 while keeping (x-5). So, I think that’s why the 2nd answer choice doesn’t work.

I think the logic of this should be - If the three numbers are 3 consecutive, then it will definitely be a multiple of 3.

1,2,3 - 1 * 2 * 3 is a multiple of 3. You can check this with any example. If you think about it, it’s actually a property of numbers. (Multiply 7 consecutive numbers - must be a multiple of 7, etc…)

But in the options, we have it in terms of x. One of the three numbers MUST be a multiple of 3 (as above) => x, or x+1, or x-1.

All options have x in them, so we need an x+1 and x-1.
But also, if x+1 is a multiple of 3, then (x+1+3), (x+1+3+3), etc will also be a multiple of 3.

In option 2, (x-5)(x)(x+4):

x-5, is equivalent to x-5+3+3, or x+1

x+4, is equivalent to x+4-3, or x+1.
So it does not have the x-1 equivalent number, so it will not be a multiple of 3. You can also try substituting 4 for x. => (4-5)(4)(4+4) = -1 * 4 * 8.

We do not need to +3 to EVERY one of the three factors simultaneously. For example, the first answer choice (which is divisible by 3) is obtained by adding 6 only to the factor $(x+1):

(x-1)x(x+1)\equiv (x-1)x(x+1+3+3) = (x-1)x(x+7)

The reason why the 2nd answer choice does not work is not because we must add or subtract 3 from every factor, but it’s because we cannot have both factors (x+4) and (x-5) at the same time in this product. As you correctly point out, we can make (x-5) by -6 from (x+1). So,

(x-1)x(x-5) is divisible by 3.

But then, we cannot make the factor (x+4) by performing \pm 3 on either (x-1) or on x.