Posted by David on Thursday, August 7, 2008 at 7:15pm.
This probably should have been written as
(2a^2+13a-7)/[(3a^3-27a)(4a^2-1)/9a^2]
Try factoring 2a^2 +13a -7 into
(2a -1)(a+7)
and 3a^3-27a into
3a(a^2-9 = 3a(a+3)(a-3)
and 4a^2 -1 into (2a +1) (2a -1)
The 3a and (2a-1) terms will cancel.
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.
What you just typed as sure as the problem as written is NOT close to what you posted. Factor, then divide out common factors.
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