http://puu.sh/3KeN2.jpg

i ended up with (x - 2)(x + 1) but that is completely off and i dont know where i went wrong...

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- Jul 23rd 2013, 06:06 PMMathnood768rational expressions #4
http://puu.sh/3KeN2.jpg

i ended up with (x - 2)(x + 1) but that is completely off and i dont know where i went wrong... - Jul 23rd 2013, 06:39 PMchiroRe: rational expressions #4
Hey Mathnood768.

Hint: x^2 - 2x - 3 = (x-3)(x+1), x^2 - x - 6 = (x-3)(x+2), x^2 + 2x = x(x+2), and x^2 - 4x = x(x-4).

Try double checking your calculations with the above results. - Jul 23rd 2013, 06:48 PMadkinsjrRe: rational expressions #4
Your major mistake is the way you go about factoring the second term. Instead of using the

*times sign x*I'm just going to write it like this:

$\displaystyle \frac{(x^2-2x-3)(x^2+2x)}{(x^2-x-6)(x^2-4x)}$

It's the same thing. You factored it all wrong, can you see where the factoring went wrong?

Just re-factor these one-by-one,

$\displaystyle (x^2-2x-3)$

$\displaystyle (x^2+2x)$

$\displaystyle (x^2-x-6)$

$\displaystyle (x^2-4x)$

Edit: I 'd like to add that all of the errors you make (that (I've seen so far) are just factoring quadratic expressions, so you should review that. You seem to know how to approach rational expressions like this methodologically (factoring, cancelling, and simplifying,) but you're hung-up on the factoring bit. - Jul 23rd 2013, 07:22 PMMathnood768Re: rational expressions #4
http://puu.sh/3KhCy.jpg

i tried again and got x - 2 ... - Jul 24th 2013, 11:31 AMMathnood768Re: rational expressions #4
i still need help

- Jul 24th 2013, 02:07 PMadkinsjrRe: rational expressions #4
It looks like you've concluded that $\displaystyle (x^2+2x)=(x+2)(x+2) $ which isn't true. $\displaystyle (x+2)(x+2)=x^2+4x+4$. When factoring just try to expand them out and ask yourself if they make any sense.

$\displaystyle (x^2-4x)=x(x-4)$ - Jul 24th 2013, 03:24 PMMathnood768Re: rational expressions #4
http://puu.sh/3KU3p.jpg

still getting it wrong