Question

What is the restriction on the quotient of quantity 3 x squared plus 12 x plus 9 divided by quantity x squared minus 4 divided by quantity 4 x plus 4 divided by quantity x plus 2?

x ≠ -2

x ≠ -1

x ≠ 1

x ≠ 2

Answer 1

@whpalmer4 @keke_luvs_u @lovatic4life @Partycool @XcoltenX @Spacelimbus @GTMclachlan @Green_Plan

Answer 2

@poopsiedoodle @Power2Knowledge @HelpBlahBlahBlah @Mr.ClayLordMath @ChelseaTheRULER @Xmoses1 @zepdrix

Answer 3

@fc3857 @nepurrta @Jonathan1997 @JA1 @secret66

Answer 4

@hba @heather040200 @hlee13 @ash2326 @austinL @Ashleyisakitty

Answer 5

Please write your equation using the equation editor button rather than making us puzzle out what it looks like...

Answer 6

ok i will hold on

Answer 7

@whpalmer4

Answer 8

i tried doing this but i kept getting answers that wasnt an answer choice

Answer 9

\[\frac{\frac{3x^2+12x+9}{x^2-4}}{\frac{4x+4}{x+2}} = \frac{3x^2+12x+9}{x^2-4}*\frac{x+2}{4x+4}\]
Restrictions on the expression are those values of \(x\) such that the denominator equals 0. Be sure to do this before you simplify, because the restrictions apply even if the terms that supply them are cancelled out during simplification!

Answer 10

What do you get when you factor the numerators and denominators of those fractions?

Answer 11

(only denominators matter for this question, but you can probably use the practice :-)

Answer 12

wouldnt the denominator be 4(x+1)

Answer 13

Well, that's the denominator for the fraction on the right. How about the one on the left?

Answer 14

ohh ummm....

Answer 15

i cant really see the fraction on the left

Answer 16

You can't see how to factor \(x^2-4\)?

Answer 17

Here's what we have after doing the division:
\[\frac{3x^2+12x+9}{x^2-4}*\frac{x+2}{4x+4}\]

Answer 18

(we converted the division into a multiplication)

Answer 19

(x-2)(x+2)

Answer 20

Very good! So the combined denominator of our multiplication will be \[4(x+1)(x-2)(x+2)\]What are the restrictions on the variable?

Answer 21

"What are the restrictions on the variable?" And from where do they stem?

Answer 22

2 right

Answer 23

Mind explaining where that came from? Is that the only restriction on x?

Answer 24

"wouldnt the denominator be 4(x+1)?" Yes, PGR! Given that this is in the denom., what can we conclude about restrictions on x?

Answer 25

Don't forget what I said earlier:

"Restrictions on the expression are those values of x such that the denominator equals 0."

Answer 26

oh i really dont know

Answer 27

but was i right its 2

Answer 28

\[4(x+1)(x-2)(x+2)=0\]Solve for \(x\)

Answer 29

You're only 1/3 right :-)

Answer 30

darn

Answer 31

For a product of terms to equal 0, one or more of the terms must equal 0, right?

So we need to solve
\[x+1=0\]\[x-2=0\]\[x+2=0\]3 different values of \(x\), all will make the denominator 0.

Answer 32

x=-1 x=2 x=-2

Answer 33

I like whpalmer's approach. He's urging you to determine every one of the factors of the denominator (there are three here), and then to set each factor = to 0, to find the restrictions on the input variable, x. (There are more complicated situations than this, but concentrate on those 3 factors in the denom. for now.)

Answer 34

but im confused what is the answer though

Answer 35

The answer is all 3 values of \(x\) that make the denominator = 0. What are they?

Answer 36

Please go back and re-read what whp has advised you to do along the way to finding those answers. Not answer, but answers (there are three). I hope you're taking notes on this discussion, as you'll need to understand and use these tactics in the future when dealing with similar problems.

Answer 37

If those values were \(x=1,x=2.x=3\), it would be written\[x\ne1,x\ne2,x\ne3\] because \(x\) is not allowed to be any of those values.

Answer 38

but thats not an answer choice

Answer 39

what I wrote was an example of how one notates a function with restrictions on the variable:

\[y = f(x), \,x\ne1, x\ne 2, x\ne3\]

Answer 40

I said "if those values were" which was the tip-off that I was not giving you the answer to your problem.

Answer 41

But why don't you tell us what the answer choices you have are...

Answer 42

There are various ways to write the SET of answers. Whereas whpalmer has written x=-1, x=2, x=-2 (which is correct), one could also write (in set notation) {-1,2,-2}.

Try to find a connection between these results and what you and whp have arrived at here.

Answer 43

i already did

Answer 44

And what did you find?

Answer 45

but im learning how how to do it but i want you to make sure that i am doing it right ok

Answer 46

I understand that, but at some point we all have to take responsibility for finding and checking our answers unaided.

So, once again, please share with us the four possible choices of answers.

Answer 47

look its at the top where you read the question...i had posted it already

Answer 48

Very good! So the combined denominator of our multiplication will be
4(x+1)(x−2)(x+2)
What are the restrictions on the variable?

Is it conceivable that you could mark / check off more than 1 of the given answer choices? As before, I believe the correct answer choices are {-1, 2, -2}.

Why don't you try to check all three of those?

Answer 49

And leave x=1 unchecked.

Answer 50

thats what i am doing right now im replacing X with those three numbers

Answer 51

OK. If you still feel stuck, I'd suggest you e-mail, call or visit your instructor, showing him/her what you have done so far.

Answer 52

thats what i am suppose to do right

Answer 53

@whpalmer4 im doing it right by replacing x with those three numbers

Answer 54

I strongly urge you to set "becoming less dependent on others for answers" as a goal. I really don't mean to be unkind, but your constant asking for confirmation does not strike me as healthy, and it makes me very uncomfortable.

I've suggested that you talk with your teacher in cases like this. Neither whpalmer nor I grade your work; your teacher does. So, would you please consider seeing your teacher if you're not satisfied with the responses/confirmation you're getting or not getting from whpalmer and me.

Answer 55

I honestly believe that whp and I have done all we can to help you with this particular problem, and that we have given it (and you) our best efforts. Please: move on.

Answer 56

i been moved what are you talking about i have already found the answer your still sitting on here writing your post here as you seen i stop looking at

Answer 57

this particular question and moved on working on my other question

Answer 58

You could have told me that earlier. Your point is what, exactly?

Answer 59

well i thought i did but now since you know im gone

Answer 60

what wsa the answer?

Answer 61

was

Answer 62

This was obviously her first time doing this, she deserved the confirmation to know if she was doing her work correctly. She seemed to actually be doing the work and learning, as some people here, do not.

Answer 63

\[x \neq 1 \]