Math
posted by David on .
No,
I'm sure the problem is written like this:
[(3a^327a)/(2a^2+13a7)]/[9a^2/4a^21)]
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^327a)/(2a^2+13a7)]/[9a^2/4a^21)]
can also be written
[(3a^327a)/(2a^2+13a7)]*[(4a^21)/9a^2]
Do the factoring of (3a^327a), (2a^2+13a7) and 4a^2 1 that I mentioned earlier. Many terms will cancel.