Question

Help wanted!

Answer 1

http://prntscr.com/9imy8o

Answer 2

@Astrophysics @ganeshie8

Answer 3

We can get rid of the first one because if we have x = 4 or 2 we get 0 right, that doesn't give us undefined. We get undefined if we are dividing by 0 right?

Answer 4

yea

Answer 5

Is there any other we can eliminate right away looking at the restrictions?

Answer 6

The second option?

Answer 7

No, we can get rid of option d because we have x-2 in the denominator, so we can't have x = 2 as it will give us 0 which means it will be undefined. But we have to realize there is another restriction which is x cannot be 4. So we can trash both options A and D.

Now looking at B and C, we'll have to factor, do you know how to factor?

Answer 8

It's ok if you don't we can work on it together

Answer 9

\[\frac{ 2 }{ x^2+6x+8 }=x^2+6x+8=x(x+2)+4(x+2)=(x+2)(x+4)=\]\[\frac{ 2 }{(x+2)(x+4)}\]

Answer 10

Nice, you factored the denominator :), so what will be the restrictions?

Answer 11

we eliminate that one as well?

Answer 12

Yes good, we can set x+2 = 0 and x+4 = 0 and solve for x which will give us our restrictions for the problem. In this case x cannot be -2 and -4.
We are left with only one option and I know you can factor!

Answer 13

lol so option c is my answer correct?

Answer 14

Or to make it more clear you can do \[x+2 \neq 0~~~\text{and}~~~x+4 \neq 0\] and solve for x so you don't get confused.

Yes, well done!

Answer 15

thanks!