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you basically just need to brush up on factoring

(a+b)^2=a^2+2ab+b^2, if you factor a 3 out of that numerator, you will have this situation

yeah how do i factor 9 to add up to 6

or do i not have to

factor out a 3 first, it will make it much easier

ive got (x+3)(x-3)

thats wrong

thats a difference of 2 squares

so what do i do

if you expanded that you'd get x^2-9

this part 3x^2+6x-9

factor out a 3

3,6,9 all have a common factor of 3 here

3(x^2 +6x-9)

what you just did will yield 3x^2 + 18x -27

3(x^2-2x-3)

+2x

good

now factor the denominator

very much the same thing, guess which number you're pulling out?

2(x^2-1)

yes

now factor (x^2-1)

(difference of 2 squares), i mentioned it a little bit ago

hmm im not sure here

hehe you just did it mistakenly earlier

remember you typed (x-3)(x+3)

oh that

(x+1)(x-1)

yes, now it's all factored. put it all together and cancel like terms

ok but could i have plug in 1 and get an answer

actually, you can further factor the numerator

right now you have the numerator at (x^2+2x-3)

you could not have p lugged 1 into the original equation before factoring it

look at the original denominator

2x^2-2

if you plugged a 1 in here, you'd get a divide by 0 error

i see you will just get 0/0

so try factoring x^2+2x-3

(x-1)(x+3)

good

now everything is factored

so when i cancel i get x+3/x+1

don't forget about the 3 and 2 you factored out

3(x+3)/(2(x+1)

at this point you can plug in 1 and you will not get a divide by 0 error

because you cancelled out the (x-1)

ok i get 12/4

when i reduce i get 3/1

yes. that's it

thanks for the help

np

i have another limit problem but its approaching infinity

ok, let's have it

x^2 + 2x-3/2x^2-2 as x approaches infinity, what is the limit

now look at this numerator and denominator

you already factored them in the last problem

give me your factored equation for this problem

(x-1)(x+3)/2(x^2-1)

the denominator further factors out to (x-1)(x+1)

yes, now cancel

x+1/x+1

or 2x+1

what happened to x+3?

my final looks like this (x+3)/(2x+1)

yes

2x+2

no

2(x+1) then you foil

that's as far as you can factor it: (x+3)/[2(x+1)]

you cant do 2x+2?

what's your thinking on that? just trying to see where you're going wrong

oh my mistake, i see what you were doing

thats not a big deal to expand the denominator to 2x+2

so youd have (x+3)/(2x+2)

i thought you meant the whole equation factored out to that

if you have 2(x+1)...why not just foil (2)(x)=2x...2(1)=2...2x+2...

you can, but it's generally not necessary after you've already factored

plus if you do it too early you'll have to factor it again later

oh ok

so since i can't plug in infinity...what should i do

now you have to think what would happen to f(x) is you kept plugging in very large numbers

it would remain large

nope

try plugging in 100 just to test it

well it might cut down...but at the end it would still be large

100+3/2(100+1)=103/202= .50

ok now 1000

the same .50

ok

so 1/2?

1 sec

that's right

ok thanks for the help

sure

just understand why that's the answer

essentially its like graphing x/(2x)

ok

thanks for the explanaition