Question

Help?

Answer 1

http://prnt.sc/a8emgy set it up like so

Answer 2

\[\frac{n-3}{n^2+6n+8}\times\frac{n+2}{n+1}\]you need to get a common denominator next, and you do this by doing something like \[\frac{(n-3)(n+1)}{(n^2+6n+8)(n+1)}\times\frac{(n+2)(n^2+6n+8)}{(n^2+6n+8)(n+1)}\]

Answer 3

dang theres more?

Answer 4

Yes haha you need to simplify XD

Answer 5

well which one of those answer choices look the most sufficient to you? bwahah

Answer 6

A lot of work, right?

Answer 7

I was debating between A AND b.

Answer 8

LOL

Let's just do this the right way XD

Answer 9

Don't be lazy amiga ;)

Answer 10

No I really dont know how to simplfy dont make me feel dumb. Im not even kidding.

Answer 11

I never meant you personally x'D I'd help you

Answer 12

BWHAAH If I knew I wouldnt be asking. promise.

Answer 13

So how shall we simplify?

Answer 14

One moment lol

Answer 15

I think I got it. !!!!!!!

Answer 16

LOL really?

Answer 17

No. A?

Answer 18

I messed up somewhere, because the numerator cannot be factored ;-;

Answer 19

Oh lord thats a lot of work. Im going to go with my gut and say its A. Ready to move to the next one?!

Answer 20

OMG...I added rather than multiplying! OMG

Answer 21

omfg.. bwahaha no you're lying...

Answer 22

Ignore all that I wrote... lol

Answer 23

\[\frac{n-3}{n^2+6n+8}\times\frac{n+2}{n+1}\]\[=\frac{(n-3)(n+2)}{(n+4)(n+2)(n+1)}\]
The n+2s cancel leaving us with \[\frac{(n-3)}{(n+4)(n+1)}~or~\frac{(n-3)}{(n^2+5n+4)}\]

Answer 24

I made it complicated because I forgot we were supposed to add rather than multiply until I looked at the original problem again >.>

Answer 25

I mean I forgot we were supposed to multiply rather than add...oh my, I'm all sorts of backwards today

Answer 26

Thank you!

Answer 27

You're so very welcome!