2a^2-4a+2/3a^2-3
Please show all work, and state any excuded values.

- anonymous

2a^2-4a+2/3a^2-3
Please show all work, and state any excuded values.

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- Luigi0210

factor

- anonymous

idk how

- Luigi0210

Well we are screwed now aren't we? :P

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## More answers

- anonymous

Ok this is what i got:
2a^2-4a+2/3a^2-3=2(a-1)/3(a+1)=\[a \neq-1\]

- anonymous

But my teacher said I need to show my work so now what? @Hero

- anonymous

@mathstudent55

- anonymous

@a

- anonymous

@anikhalder @abbiebosworth @aditkarekatte @ariannafar1 @alli14344 @Ashleyisakitty @bobsr. @countrygirl33 @dumbcow @dan815 @deadman340

- Luigi0210

@Jhannybean she'll take care of you

- Jhannybean

omg lol he tagged.....a lot of people.

- anonymous

lol

- anikhalder

the second a^2 in 2/3a^2 is in the denominator?

- Jhannybean

\[\large \frac{2a^2-4a+2}{3a^2-3}\] first step, let's start with the numerator.
divide the numerator by 2. what do you get?

- Luigi0210

wait for it...

- Jhannybean

second step, divide denominator by 3, what do you get?

- anonymous

Um idk 3a^2?

- Luigi0210

\[\frac{ 2(a^2-2a+1) }{ 3(a^2-1) }\]

- Jhannybean

lol not factor OUT, divide!!!

- Luigi0210

I'm not sure if you can divide..

- Luigi0210

but then again you're the expert

- Jhannybean

numerator : (2a^2)/2 - (4a)/2 + ( 2 )/2
denominator (3a^2)/3 - (3)/3

- Jhannybean

can you simplify these? lol. That's what i meant.

- anonymous

ok
2(a-1)/3(a+1)?

- Luigi0210

see you can factor, good job

- Luigi0210

forgot to factor that 3 Jhann

- anonymous

yea but thts wht i did and then i put a doesnt equal -1 but my teacher said i need to show my work...WTF

- Jhannybean

OMGGGG..

- anonymous

|dw:1370420792468:dw|

- Luigi0210

Show that there is a hole

- Jhannybean

\[\large \frac{2a^2-4a+2}{3a^2-3}= \frac{a^2-2a+1}{a^2-1}\]

- anonymous

ok so should i write that down?

- Luigi0210

Haha, Oh Jhan :)

- Jhannybean

-_-...

- Luigi0210

Say that it can't equal one because that will the make the function undefined

- Jhannybean

so what two numbers multiply to give +1 but add to give -2?

- Luigi0210

Kinda late, he solved it already :P

- Jhannybean

oh lol. nvm :D good job.

- anonymous

- Jhannybean

(2a - 2)(a-1) = 0
(2a - 2) = 0
a = 1.
a - 1= 0
a = 1

- Jhannybean

so what are you trying to show exactly...

- Luigi0210

Here's a pretty graph.. hm

##### 1 Attachment

- anonymous

1sec..ok ill attach a document

- Jhannybean

\[\frac{a^2-2a+1}{a^2-1} = \frac{\cancel{(a-1)}(a-1)}{\cancel{(a-1)}(a+1)} = \frac{(a-1)}{(a+1)} \therefore a \neq -1\]

- Jhannybean

@Luigi0210 how do you do those graphsss

- anonymous

ok it's this

##### 1 Attachment

- Jhannybean

@Pyromancer follow what i just did :) that's how you show there is a vertical asymptote at a=-1.

- anonymous

|dw:1370421280822:dw|

- Luigi0210

It's an ipod app.. pretty cool if you ask me

- anonymous

after factorization

- Jhannybean

you guys know you can CANCEL our like terms???? -_- Like i showed in the beginning....lol Omg.

- anonymous

So what should I writ down for number seven?

- Jhannybean

at a=1 you have a horizontal asymptote, and at a=-1 you have a vertical asymptote.

- Luigi0210

apparently we like taking the long challenging road x)

- anonymous

|dw:1370421488650:dw|

- anonymous

ok but work wis, just like the original problem then..

- anonymous

for upper part i did factorized in earlier image thhe fotstr one.

- anonymous

@Jhannybean you cant eleiminate 2/3.

- Luigi0210

nor divide I think

- Jhannybean

@deadman340 you can eliminate common multiples in the numerator and denominator separately.

- anonymous

so you did it.

- anonymous

|dw:1370421625220:dw|

- Jhannybean

ok nevermind you guys are right.you can only factor stuff out :( so you will have a (2/3) constant.

- Jhannybean

Haha. I thought you could. :|

- anonymous

thts wht i did b4 but my teacher said i need to show my work -.-

- anonymous

(adding to the dra

- anonymous

|dw:1370421737897:dw|

- anonymous

...wing)

- anonymous

it is 100% right what i did.

- Jhannybean

yes you are right @deadman340 .

- anonymous

now u are doing something else you are not sure youself what are you doing

- Jhannybean

what are you talking about? Iam correcting my mistake...

- anonymous

in first you just eliminated the 2/3 now you did right.

- Luigi0210

btw lost your 3

- Jhannybean

ok/ and that means i'm unsure of what i'm doing? I'm correcting my mistake, that doesn't make me wrong.

- anonymous

good you have done right is last.

- anonymous

dont worry both of our methods are right.

- anonymous

Ok let me scan wht i hav so far...

- Jhannybean

\[\large \frac{2a^2-4a+2}{3a^2-3}= \frac{2(a^2-2a+1)}{3(a^2-1)}\]\[\large \frac{2(a^2-2a+1)}{3(a^2-1)}= \frac{2(a-1)(a-1)}{3(a-1)(a+1)}= \frac{2(a-1)}{3(a+1)}\]

- anonymous

hey your factorization for 2(a^2-2a+1) is wrong it dont produces (a-1)(a-1)

- anonymous

ok look this over..PUHLEAZEEE

##### 1 Attachment

- Luigi0210

Yea it does

- Jhannybean

that's what i have?....

- anonymous

so thts good?

- anonymous

oh sorry that produce that my mistake.

- anonymous

bro both mehods are right its your choice .

- anonymous

ok..one more q?

- anonymous

after i close this... lol

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