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

Algebra  K, Monday, April 27, 2009 at 7:21pm
Wait, I wrote it wrong. This is the correct way:
(9z42z72)/(3z^2+28z+32)divided by(z^29z+18)/(2z^2+10z48)

Algebra  Reiny, Monday, April 27, 2009 at 7:33pm
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

Algebra  K, Monday, April 27, 2009 at 7:43pm
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.

Algebra  Reiny, Monday, April 27, 2009 at 7:47pm
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

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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