#16h^2+8hk-15k^2# is homogeneous: all the terms are of order 2. As a result, factoring this polynomial is similar to the problem of factoring the polynomial #16x^2+8x-15#, which is of the form #ax^2+bx+c#, with #a=16#, #b=8# and #c=-15#.
This has discriminant given by the formula:
#Delta = b^2-4ac#
#= 8^2-(4xx16xx-15) = 64+960 = 1024 = 32^2#
So #16x^2+8x-15=0# has roots given by:
#x = (-b+-sqrt(Delta))/(2a) = (-8+-32)/32 = (-1+-4)/4#
That is #x = -5/4# or #x = 3/4#.
Multiplying these two equations through by the denominators, we can deduce that #(4x+5)# and #(4x-3)# are factors:
Hence #16x^2+8x-15 = (4x+5)(4x-3)#
If we substitute #h/k# for #x#, this becomes:
#16(h/k)^2+8(h/k)-15 = (4(h/k)+5)(4(h/k)-3)#
Now multiply both sides by #k^2# to find
#16h^2+8hk-15k^2 = (4h+5k)(4h-3k)#
