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)?
Many thanks in advance!
I think this is what you’re saying:
(x-1)x(x+1) \equiv (x-1 + 3)x(x+1 - 3 -3) = (x+2)x(x-5)
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.
Just to confirm, if I choose to add 3 to one of the brackets, then I must do the same for the others?
Thank you sooo much for your explanation btw!
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):
Hi!
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.
I hope this helps!
Yes - check your computation of (x - 4)(x)(x + 10).
