By the polynomial remainder theorem, 
will be a factor of 
if the remainder upon division is 0, and this remainder is given by 
:
Since
, it follows that 
. But notice that if 
, then we have
and since
, the equation above requires that either 
or 
, which is impossible. So 
.
So we have 8 cases to check:
(1) Notice that if
, we have 
. This is only possible for 
.
(2) If
, then 
, and so we can have 
.
(3) If
, then 
, and so 
.
(4) If
, then 
.
(5) If
, then 
.
(6) If
, then 
.
(7) If
, then 
.
(8) If
, then 
.
At the same time, we have 8 cases to consider to find how many options there are for
.
(1). We have only one choice of 
.
(2). This is the same as when , which we found to be 2 choices.
(3) Same as; 3 choices.
(4) Same as; 4 choices.
(5) 5.
(6) 4.
(7) 3.
(8) 2.
In total, there are
ways to choose
such that is a factor of 
, so the answer is B.
Note the symmetry of the sum above. You can easily give a slightly briefer combinatorial argument for this answer, but I figured a more brute-force approach would be easier to follow.
Since
and since
So we have 8 cases to check:
(1) Notice that if
(2) If
(3) If
(4) If
(5) If
(6) If
(7) If
(8) If
At the same time, we have 8 cases to consider to find how many options there are for
(1)
(2)
(3) Same as
(4) Same as
(5) 5.
(6) 4.
(7) 3.
(8) 2.
In total, there are
ways to choose
Note the symmetry of the sum above. You can easily give a slightly briefer combinatorial argument for this answer, but I figured a more brute-force approach would be easier to follow.