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September 17, 2014

September 17, 2014

Posted by **David** on Thursday, August 7, 2008 at 7:51pm.

I'm sure the problem is written like this:

[(3a^3-27a)/(2a^2+13a-7)]/[9a^2/4a^2-1)]

And to divide you multiply by the reciprocal.

So i still don't get the problem I also know to factor, but I'm stil confused.

- Math -
**drwls**, Thursday, August 7, 2008 at 8:47pm[(3a^3-27a)/(2a^2+13a-7)]/[9a^2/4a^2-1)]

can also be written

[(3a^3-27a)/(2a^2+13a-7)]*[(4a^2-1)/9a^2]

Do the factoring of (3a^3-27a), (2a^2+13a-7) and 4a^2 -1 that I mentioned earlier. Many terms will cancel.

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