We can simply check the divisors of the given X and count how many of those divisors multiplied by each other gets X.
For the i-th divisor we check all the previous j-th divisor, for 0 <= j <= i. When divisor_i * divisor_j == X is false for every j in the on going iteration, we can return the current count since divisor_i will be smaller in the next iteration and every upcoming check of divisor_i * divisor_j == X will be false.
The running time in the worst case is O(n^2) for n being the length of the matrix.