Asked by David

I'm totally stuck on this problem!!

Simplify:
(2a^2+13a-7/3a^3-27a)(4a^2-1/9a^2)
Answered by drwls
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.
Answered by David
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.
Answered by bobpursley
What you just typed as sure as the problem as written is NOT close to what you posted. Factor, then divide out common factors.
There are no AI answers yet. The ability to request AI answers is coming soon!