In how many ways can 2700 be written as a product of 8 distinct integers?

2700 = $$2^2 * 3^3 * 5^2$$ = 1*2*2*3*3*3*5*5
These are 8 numbers (not distinct). If we combine any two of them, the number becomes less than 8.
So, the number of ways of writing 2700 as a product of 8 distinct integers is 0.

And would be 6 as we can also take into account the negative numbers.

Integers include negative numbers. so why can't I take the other number as a negtive and solve?