Factor again the expression

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- pokemon23

Factor again the expression

- schrodinger

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- pokemon23

|dw:1328231408045:dw|

- pokemon23

would the answer be

- anonymous

they both have x in common so...

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## More answers

- pokemon23

|dw:1328231433797:dw|

- anonymous

yep!

- pokemon23

yes I'm getting the hang of it

- pokemon23

thanks meidrik and Hero

- anonymous

sure thing

- pokemon23

|dw:1328231509108:dw|

- Hero

What did I do? I didn't do anything :P

- anonymous

It was your presence

- Hero

All I did was show up :P

- pokemon23

do we find facor in this question for this question

- anonymous

they have a 6 in common so take that out

- pokemon23

|dw:1328231568890:dw|

- anonymous

6(2x-r)

- Hero

You find something that you can divide by both terms evenly

- pokemon23

I don't get it,,?

- pokemon23

So we divide 12 by 3 or 2..?

- anonymous

by 6

- anonymous

you divide by what you have in common

- pokemon23

|dw:1328231685504:dw|

- anonymous

right, but it isnt simplified because they still have a 3 in common

- pokemon23

|dw:1328231766372:dw|

- Hero

|dw:1328231645628:dw|

- anonymous

no, 2(6x - 3r), factor a 6 OUT
(2)(6)(2x - r) = 12(2x-r)

- Hero

mridrik, I was referring to the previous problem, not this one.

- anonymous

I was talking to the other guy

- Hero

oops

- pokemon23

lost?

- Hero

bad timing on your post

- anonymous

yeah

- anonymous

(Talking to pokemon)
still stuck?

- pokemon23

let me relog out then reconnect..

- pokemon23

ok back I'm stuck lets go step by step

- anonymous

so which one are we on again?

- pokemon23

|dw:1328232239409:dw|

- anonymous

They are both divisible by 6, so divide by 6
6(2x - r)

- pokemon23

would it be 2t

- anonymous

look at what i wrote

- pokemon23

ya but in this question the variable consist of a t

- anonymous

I thought it was an x

- pokemon23

lol go to my next question I posted

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