anonymous
  • anonymous
2a^2-4a+2/3a^2-3 Please show all work, and state any excuded values.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Luigi0210
  • Luigi0210
factor
anonymous
  • anonymous
idk how
Luigi0210
  • Luigi0210
Well we are screwed now aren't we? :P

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More answers

anonymous
  • anonymous
Ok this is what i got: 2a^2-4a+2/3a^2-3=2(a-1)/3(a+1)=\[a \neq-1\]
anonymous
  • anonymous
But my teacher said I need to show my work so now what? @Hero
anonymous
  • anonymous
@mathstudent55
anonymous
  • anonymous
@a
anonymous
  • anonymous
@anikhalder @abbiebosworth @aditkarekatte @ariannafar1 @alli14344 @Ashleyisakitty @bobsr. @countrygirl33 @dumbcow @dan815 @deadman340
Luigi0210
  • Luigi0210
@Jhannybean she'll take care of you
Jhannybean
  • Jhannybean
omg lol he tagged.....a lot of people.
anonymous
  • anonymous
lol
anikhalder
  • anikhalder
the second a^2 in 2/3a^2 is in the denominator?
Jhannybean
  • 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
  • Luigi0210
wait for it...
Jhannybean
  • Jhannybean
second step, divide denominator by 3, what do you get?
anonymous
  • anonymous
Um idk 3a^2?
Luigi0210
  • Luigi0210
\[\frac{ 2(a^2-2a+1) }{ 3(a^2-1) }\]
Jhannybean
  • Jhannybean
lol not factor OUT, divide!!!
Luigi0210
  • Luigi0210
I'm not sure if you can divide..
Luigi0210
  • Luigi0210
but then again you're the expert
Jhannybean
  • Jhannybean
numerator : (2a^2)/2 - (4a)/2 + ( 2 )/2 denominator (3a^2)/3 - (3)/3
Jhannybean
  • Jhannybean
can you simplify these? lol. That's what i meant.
anonymous
  • anonymous
ok 2(a-1)/3(a+1)?
Luigi0210
  • Luigi0210
see you can factor, good job
Luigi0210
  • Luigi0210
forgot to factor that 3 Jhann
anonymous
  • 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
  • Jhannybean
OMGGGG..
anonymous
  • anonymous
|dw:1370420792468:dw|
Luigi0210
  • Luigi0210
Show that there is a hole
Jhannybean
  • Jhannybean
\[\large \frac{2a^2-4a+2}{3a^2-3}= \frac{a^2-2a+1}{a^2-1}\]
anonymous
  • anonymous
ok so should i write that down?
Luigi0210
  • Luigi0210
Haha, Oh Jhan :)
Jhannybean
  • Jhannybean
-_-...
Luigi0210
  • Luigi0210
Say that it can't equal one because that will the make the function undefined
Jhannybean
  • Jhannybean
so what two numbers multiply to give +1 but add to give -2?
Luigi0210
  • Luigi0210
Kinda late, he solved it already :P
Jhannybean
  • Jhannybean
oh lol. nvm :D good job.
anonymous
  • 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
  • Jhannybean
(2a - 2)(a-1) = 0 (2a - 2) = 0 a = 1. a - 1= 0 a = 1
Jhannybean
  • Jhannybean
so what are you trying to show exactly...
Luigi0210
  • Luigi0210
Here's a pretty graph.. hm
1 Attachment
anonymous
  • anonymous
1sec..ok ill attach a document
Jhannybean
  • 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
  • Jhannybean
@Luigi0210 how do you do those graphsss
anonymous
  • anonymous
ok it's this
Jhannybean
  • Jhannybean
@Pyromancer follow what i just did :) that's how you show there is a vertical asymptote at a=-1.
anonymous
  • anonymous
|dw:1370421280822:dw|
Luigi0210
  • Luigi0210
It's an ipod app.. pretty cool if you ask me
anonymous
  • anonymous
after factorization
Jhannybean
  • Jhannybean
you guys know you can CANCEL our like terms???? -_- Like i showed in the beginning....lol Omg.
anonymous
  • anonymous
So what should I writ down for number seven?
Jhannybean
  • Jhannybean
at a=1 you have a horizontal asymptote, and at a=-1 you have a vertical asymptote.
Luigi0210
  • Luigi0210
apparently we like taking the long challenging road x)
anonymous
  • anonymous
|dw:1370421488650:dw|
anonymous
  • anonymous
ok but work wis, just like the original problem then..
anonymous
  • anonymous
for upper part i did factorized in earlier image thhe fotstr one.
anonymous
  • anonymous
@Jhannybean you cant eleiminate 2/3.
Luigi0210
  • Luigi0210
nor divide I think
Jhannybean
  • Jhannybean
@deadman340 you can eliminate common multiples in the numerator and denominator separately.
anonymous
  • anonymous
so you did it.
anonymous
  • anonymous
|dw:1370421625220:dw|
Jhannybean
  • Jhannybean
ok nevermind you guys are right.you can only factor stuff out :( so you will have a (2/3) constant.
Jhannybean
  • Jhannybean
Haha. I thought you could. :|
anonymous
  • anonymous
thts wht i did b4 but my teacher said i need to show my work -.-
anonymous
  • anonymous
(adding to the dra
anonymous
  • anonymous
|dw:1370421737897:dw|
anonymous
  • anonymous
...wing)
anonymous
  • anonymous
it is 100% right what i did.
Jhannybean
  • Jhannybean
yes you are right @deadman340 .
anonymous
  • anonymous
now u are doing something else you are not sure youself what are you doing
Jhannybean
  • Jhannybean
what are you talking about? Iam correcting my mistake...
anonymous
  • anonymous
in first you just eliminated the 2/3 now you did right.
Luigi0210
  • Luigi0210
btw lost your 3
Jhannybean
  • 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
  • anonymous
good you have done right is last.
anonymous
  • anonymous
dont worry both of our methods are right.
anonymous
  • anonymous
Ok let me scan wht i hav so far...
Jhannybean
  • 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
  • anonymous
hey your factorization for 2(a^2-2a+1) is wrong it dont produces (a-1)(a-1)
anonymous
  • anonymous
ok look this over..PUHLEAZEEE
Luigi0210
  • Luigi0210
Yea it does
Jhannybean
  • Jhannybean
that's what i have?....
anonymous
  • anonymous
so thts good?
anonymous
  • anonymous
oh sorry that produce that my mistake.
anonymous
  • anonymous
bro both mehods are right its your choice .
anonymous
  • anonymous
ok..one more q?
anonymous
  • anonymous
after i close this... lol

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