Find the number of ways in which 8064 can be resolved as the product of two factors?

i think if we can take out the number of factors of 8064 which is (2^7)x(3^2)x(7^1)
and we know that N = Xa × Yb × Zc then
Total Number of Factors for N = (a+1) (b+1) (c+1)
therefore for 8064 =(7+1)(2+1)(1+1) => 8*3*2 => 48 number of different factors
so these 48 FACTORS include 1,2,3,4,6.......all the factor and if we divide it by 2 then we can say that all those different 48 factors are paired in pair of two and only those two pair whose product will form 8064.
so we will get 24 as our answer.
if you didn't understand then take the example given below:

for N=90, Total Number of Factors for N =12
1,2,3,5,6,9,10,15,18,30,45,90.
so obviously we can say that there are 6 pairs which are: (1,90),(2,45),(3,30),(5,18),(6,15),(9,10)

Here N=8064=(2^7)×(3^2)×(7^1) is not a perfect square so answer is (1/2)×[(7+1)*(2+1)*(1+1)]=24

If N is not a perfect square then the number of ways of putting N as a product of two natural numbers is (1/2)×[(a1+1)×(a2+1)×...×(ak+1)]
Where
N=(p1^a1)×(p2^a2)×....×(pk^ak) , p_i s are primes

◆if N is a perfect square then the result is
is (1/2)×[(a1+1)×(a2+1)×...×(ak+1)+1]

sir,
Any easy method to solve this problem

give me the simple way other than this