Posted by **David** on Thursday, August 7, 2008 at 7:15pm.

I'm totally stuck on this problem!!

Simplify:

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

- Math -
**drwls**, Thursday, August 7, 2008 at 7:36pm
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.

- Math -
**David**, Thursday, August 7, 2008 at 7:49pm
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.

- Math -
**bobpursley**, Thursday, August 7, 2008 at 7:53pm
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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