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@DariusX Do you think you could help me with this one? I would really appreciate it.
I don't understand.
I think the girl might be correct because don't the opposite things cancel each other out?
i'm only gonna do the top side, you will do the bottom
Okay, so the bottom would be |dw:1449522084926:dw|?
no, not really (x+6)(x-3) = x^2 - 3x + 6x -18 it is also very important to look at the sign before the number in this case you multiply like this: x*x + x*(-3) + 6*x + 6*(-18) = x^2 - 3x + 6x - 18 = x^2 + 3x -18
So, the bottom to your second part is: |dw:1449522644513:dw|
not your second part i mean is this the correction?
yes, now it's correct
This is answer for the top |dw:1449522899923:dw| now follow these instructions and solve for the bottom
of course it's not final, you have to subtract 4x^2 - 3x^2 = x^2 and -12x-18x = -30x but you probably know that
Okay. Give me one sec.
i'm confused. are the signs going to be minus, plus, minus, again?
no and don't worry there's nothing to be confused about it's very simple if you multiply let's say: -3x(5x - 2) it will be like this: -3x*5x + (-3)*(-2) = -15x + 6 do you understand now? you sum every multiplication, but when multiplying you multiply not only the numbers, but also their signs which are IN FRONT of them.
i see it didn't show the full picture. the top was final answer already multiplied with (x+4) and there is also not -18, but -18x
right! i know, sorry. I have been doing math all day and my brain is not functioning as efficiently as i would like.
You don't need to expand anything here
just look at the terms in the denominator of your rational function
what will the value of this expression be when \(x=-6\)?
I think i just have to justify who is correct for this problem. I don't think i have to do all of this work that I'm drawing out. I just have to explain, like a summary, of why Jamal or Angie is correct. Give me one sec @asnaseer, let me see.
oh yeah, sorry, didn't read the top. when x = –6 or x = 3 or x = 4 the denominator ends up 0 and it can never be 0 if a denominator is 0 then the function is undefined
and when the denominator is zero then you are effectively dividing by zero which is undefined
yes. so is Angie correct?
yes, Angie is correct. at those x values function is undefined
Great thank you so much! I wish I could give you both a medal. How about you answer my next "hard" question and I'll give one of you a medal for that? I'll post in right now.
I am on this site more to help students rather than to collect medals - but thanks any way :)