I think this problem was botched. The other tutor's answer in incorrect because
you don't have a factor if the remainder isn't 0. And certainly a remainder of
100 is not a remainder of 0. 

I had to change your last term +8 to +108, because otherwise none of those,
not even x-12, would have been factors. So I assumed the last term should have
been +108, not +8.

64x³-768x²-9x+108; x-12 

12|64 -768 -9  108
  |    768  0 -108
   64    0 -9    0

So 64x³-768x²-9x+108 factors as  

(x-12)(64x²-9)

Then 64x²-9 factors as (8x-3)(8x+3)

So the complete factorization is

(x-12)(8x-3)(8x+3)

So all 7 factors are:

1.  1
2.  x-12 
3.  8x-3 
4.  8x+3 
5.  (x-12)(8x-3) = 8x²-99x+36 
6.  (x-12)(8x+3) = 8x²-93x-36
7.  (8x-3)(8x+3) = 64x²-9 

On your list, A, B, and C are factors, but D is not a factor. 
(Incidentally, B and C are the same.)

Edwin