No,
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.
[(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.