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When is the bottom 0?

For what values of x make the bottom 0?

not sure?

Do you know how to factor?

Or even use the quadratic formula to solve a quadratic equation?

You can even do completing the square.

so x^2+12x+27=0?

how do i do that?

Do you have -12x or +12x?

its postive 12x

In other words find two factors of 27 that add up to be 12.

its like this |dw:1370027288868:dw|

err so it is -12x?

noo x^2 minus 12x

I'm really confused what problem we are doing because twice now you have written +12x

You have also said twice that we have -12x

sorry

this is the problem were doing

so which numbers do we use for x-m and x-n?

Have you found factors of 27 that have product 27 and sum -12?

Those are your m and n.

c is 27 and b is -12

|dw:1370027980807:dw|

See how I factored 27?

1(27)=27
3(9)=27
-1(-27)=27
-3(-9)=27

Yes I see on my answes I have
x= -9, x =-3 is that right?

Do any of those factors above that multiplied to be 27 also add up to be -12?

are these numbers removable?

and -3?

(x-3)(x-9)=0
Set both equal to 0.

x-3=0 or x-9=0
I leaving solving these linear equations to you.

do we also use -1 or no?

Where does that come from?

Have you solve x-3=0 or x-9=0 yet?

no idk how

x-3=0
You can't solve this for x?
You don't know how to get x by itself?

To undo subtraction by three you add three.

Try adding three to both sides and tell me what the result is.

x-3=0
add 3 to both sides
+3 +3
---------
what is the result?

x-3+3=0+3

what is -3+3?
what is 0+3?

0 and 3

yes so
x-3+3=0+3
gives us
x+0=3
Or just x=3

Now you solve x-9=0

What are you going to add to both sides?

9?

yes so x=?

9 and 0?

so x=9

yes

Discontinuous at x=3 or x=9

Now go back to the problem where you had everything factored.

\[\frac{x-3}{(x-3)(x-9)}\]

This was the problem where everything is factored, correct?

Does anything cancel? If so that is where you have your removable discontinuity.

This is just like matching socks.

I have to go to the grocery store. Good luck.

Thanks for all ur help