Question

How do u factor this?

Answer 1

This is also for limits

Answer 2

well you want to get rid of the compound fraction
multiply top and bottom by 3(3+x)
this is the lcm of the bottoms of the mini-fractions

Answer 3

lcm? means highest degree?

Answer 4

least common multiple

Answer 5

least common uhm

Answer 6

oh

Answer 7

So do i multiply the 1's too?

Answer 8

you just multiply top and bottom by 3(3+x) like this:
\[\frac{x}{\frac{1}{3+x}-\frac{1}{3}} \cdot \frac{3(3+x)}{3(x+3)} =\frac{x \cdot 3(3+x)}{(\frac{1}{3+x}-\frac{1}{3}) \cdot 3(x+3)} \\ =\frac{3x(3+x)}{\frac{1}{3+x} \cdot 3(x+3)-\frac{1}{3} \cdot 3(x+3)} \text{ by distributive property } \\ =\frac{3x(3+x)}{\frac{3(x+3)}{3+x}-\frac{3(x+3)}{3}}\]

Answer 9

wow how did you do it so fast it was taking me or ever to type

Answer 10

Ok uhm

Answer 11

i'm a computer

Answer 12

we are programmed to be faster

Answer 13

anyways I hope you see why we did this

Answer 14

Oh wow so you typedit all out?

Answer 15

(x+3)/(x+3)=1
and 3/3=1

Answer 16

Ok yes i see it so now i cross them out?

Answer 17

yes i did

Answer 18

yes

Answer 19

you will compound fraction go bye bye

Answer 20

i got 3-x+3 on the bottom after corssing about bottom fractions

Answer 21

well that isn't totally right
I think you should have gotten 3-(x+3)

Answer 22

oh yes

Answer 23

3-x-3

Answer 24

-x

Answer 25

\[\frac{3x(3+x)}{3-(x+3)}=\frac{3x(3+x)}{-x}\]

Answer 26

one more thing can happen

Answer 27

ok awesome so i can uhm cross out the x?

Answer 28

yes because I assume you are taking the limit to x=0 (which means x cannot be zero so we can cancel out the x/x part)

Answer 29

Yes, holy crap how did you know

Answer 30

i need to learn math

Answer 31

lol you said it had something to do with limits
and we couldn't plug in 0 at first because we had issue

Answer 32

Oh because there is a negative

Answer 33

oh yes haha good thinking

Answer 34

Ok so we cant crouss out the 0?

Answer 35

the 0?

Answer 36

Hi Ganshie

Answer 37

I mean the X

Answer 38

since x approaches 0 (this means x isn't 0 we only approach 0)
so yes you can cross out the x's
that is you can use x/x is 1

Answer 39

you should get
-3(3+x)

Answer 40

as x approaches 0

Answer 41

wait but i thought we cross out the X

Answer 42

so now we can direct sub since this function is continuous at x=0

Answer 43

oh nvm

Answer 44

we did

Answer 45

so 9+x/-1

Answer 46

\[\frac{3x(3+x)}{3-(x+3)}=\frac{3\cancel{x}(3+x)}{-\cancel{x}} \]

Answer 47

9+3x/-1

Answer 48

3(3+x)=9+3x

so you have (9+3x)/-1

Answer 49

right on

Answer 50

Thanks

Answer 51

with the ( ) of course but yeah i know what you meant

Answer 52

Yeah, lol Wait do you know how to program C++?

Answer 53

no i was not a computer
if i was i bet i could c++
I lied I'm sorry :p

Answer 54

Lol, your not a computer programmer?

Answer 55

no

Answer 56

nor a computer

Answer 57

oh lol but anyways your still fast , lol yes i know your not a computer

Answer 58

I wish you did you could help me with C++!

Answer 59

lol haha i thought i fooled ya

Answer 60

Lol i woulda been like your not?!?! Ok so what now with tihs 9+3x/-1

Answer 61

you can also write it as -(9+3x)

Answer 62

Oh ok

Answer 63

or write it as -9-3x

Answer 64

So do i just plug any number to X? until it reaches 0?

Answer 65

but remember if you do as x approaches 0 you still need to plug in 0

Answer 66

So the answer is -9

Answer 67

Woo awesome i was trying to figre out how to get there

Answer 68

\[\lim_{x \rightarrow 0} \frac{x}{\frac{1}{3+x}-\frac{1}{3}}=\lim_{x \rightarrow 0}(-9-3x)=-9-3(0)\]

Answer 69

Lol ok thanks! I have a test tommorow and i gottal earn them all today -sad-

Answer 70

good luck on your test and your learning

Answer 71

Thanks !!