Posted by **K** on Monday, April 27, 2009 at 7:19pm.

9z-42z-72/3z^2+28z+32divided by 2z^2+10z-48/z^2-9z+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*2z-6/z-3

- Algebra -
**K**, Monday, April 27, 2009 at 7:21pm
Wait, I wrote it wrong. This is the correct way:

(9z-42z-72)/(3z^2+28z+32)divided by(z^2-9z+18)/(2z^2+10z-48)

- Algebra -
**Reiny**, Monday, April 27, 2009 at 7:33pm
my factors were

[(3z+4)(3z-18)]/[3z+4)(z+8)] x [(2z-6)(z+8)]/[(z-3)(z-6)]

= 6 , z not equal to -4/3, -8, 3, or 6

- Algebra -
**K**, Monday, April 27, 2009 at 7:43pm
I don't understand. If you do it that way, your z-3 doesn't cancel and then i don't know where to go from there.

- Algebra -
**Reiny**, Monday, April 27, 2009 at 7:47pm
notice at the top there was a 2z-6

which is 2(z-3)

there is also 3z-18 at the top which

is 3(z-6)

so the z-3 and the z-6 at the bottom cancel with the rest of the stuff leaving only the 3x2 at the top

- Algebra -
**K**, Monday, April 27, 2009 at 8:08pm
Thank you for answering, but I'm so confused. Can someone else help, please?

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