Answer:
(2z-5)(2z-3)
if the "A" term in a quadratic Ax^2 + Bx + C is NOT a "1"
first check to see if there is a commpon factor of a,b,c that can be pulled out
for example 2x^2 + 4x + 6 you can pull
out a 2.... 2(x^2 + 2x +3)
if not you multiply the values a*c
in you case 4*15 which is 60
now you factor the 60
1*60
2*30
-1*-60
-2*-30
10*6
-10*-1
etc
when you find the combination that ADDS UP TO THE B TERM
you rewrite the B term using the factors you forud...
SO -16z becomes -10z - 6z
put it in the original problem
4z^2 -10z - 6z + 15
if you get this far , the GROUPS WILL ALWAYS WORK
pull out a 2z from "4z^2 -10z"
and pull out a -3 from "- 6z + 15"
and magically you will see that the "left over stuff" is the same
2z(2z-5) -3(2z-5)
pull out common item
(2z-5) (2z-3) ... all done
Explanation:
multiple AC ... 15*4 = 60
factor 60 to end up with -16
-10 * -6 = 60
rewrite (factor by grouping)
4z^2 -10z - 6z + 15
find groups
2z(2z-5) -3(2z-5)
(2z-5)(2z-3)