Question

16z^2/2z^2-8z+8*2z-4/4z

Answer 1

8z^4+7z

Answer 2

far from what I got! what did you do? please

Answer 3

16z^2 = (4z)(4z), so one 4z up top cancels with the 4z down below


2z^2-8z+8 = 2(z-2)(z-2) and
2z-4 = 2(z-2)

So the 2's outside the parenthesis cancel and one copy of 'z-2' cancels


This leaves you with \[\large \frac{4z}{z-2}\]

Answer 4

i must have copied incorrectly

Answer 5

it's \[\large \frac{16z^2}{2z^2-8z+8}\times\frac{2z-4}{4z}\]

Answer 6

oooooooooooh ok sorry

Answer 7

yeah I had to look the problem 100 times over to make sure lol

Answer 8

thanks! ;)