# I got stuck...mmm...I keep thinking how to get out here

• Mar 19th 2009, 08:12 AM
Alienis Back
I got stuck...mmm...I keep thinking how to get out here
See attached file, please
• Mar 19th 2009, 09:28 AM
oswaldo
2
(6a – 10) (a + 2) (10a – 2) (a – 2)
(2a + 4) (2a – 4) (3a – 5) (9a2 + 15a + 25)
2

must cancel (2a + 4) not (2a - 4). Be careful.
-O
• Mar 19th 2009, 09:34 AM
masters
Quote:

Originally Posted by Alienis Back
See attached file, please

Hi Alienis,

$\frac{6a^2+2a+20}{4a^2-16}\cdot \frac{10a^2-22a+4}{27a^3-125}$

I'm going to approach this from a different angle. First factor out any common factors in your numerators and denominators.

$\frac{2(3a^2+a+10}{4(a^2-4)}\cdot \frac{2(5a^2-11a+2)}{27a^3-125}$

Next, factor what you have left.

$\frac{2(3a-5)(a+2)}{4(a-2)(a+2)}\cdot \frac{2(5a-1)(a-2)}{(3a-5)(9a^2+15a+25)}$

Now, simplify all you can

$\frac{\rlap{---}2(\rlap{---------}3a-5) \rlap{----------}(a+2)}{\rlap{---}4\rlap{----------}(a-2) \rlap{----------}(a+2)}\cdot \frac{\rlap{---}2(5a-1) \rlap{----------}(a-2)}{\rlap{----------}(3a-5)(9a^2+15a+25)}$

$\frac{5a-1}{9a^2+15a+25}$