Algebra
posted by K on .
9z42z72/3z^2+28z+32divided by 2z^2+10z48/z^29z+18
Can someone help me? I know they're supposed to be flipped and factored, and I did that, but afterwards I can only get as far as:
9z+12/3z+4*2z6/z3

Wait, I wrote it wrong. This is the correct way:
(9z42z72)/(3z^2+28z+32)divided by(z^29z+18)/(2z^2+10z48) 
my factors were
[(3z+4)(3z18)]/[3z+4)(z+8)] x [(2z6)(z+8)]/[(z3)(z6)]
= 6 , z not equal to 4/3, 8, 3, or 6 
I don't understand. If you do it that way, your z3 doesn't cancel and then i don't know where to go from there.

notice at the top there was a 2z6
which is 2(z3)
there is also 3z18 at the top which
is 3(z6)
so the z3 and the z6 at the bottom cancel with the rest of the stuff leaving only the 3x2 at the top 
Thank you for answering, but I'm so confused. Can someone else help, please?