Question

Please Help!!

Answer 1

what do you need help with?

Answer 2

just posted it sorry

Answer 3

okay, so you first need to factor all the equations, did you do that?

Answer 4

no not yet but which equations

Answer 5

you need to factor both sides so the first equation looks like this right now
\[x ^{2}+3x+2\]Can you factor it?

Answer 6

im not very good at factoring

Answer 7

remember in order to factor this equation you must find the two numbers that multiply to get 2 and add to get three.
So you'll set it up like this
\[(x+?)\times(x+?)=x ^{2}+3x+2\]

Answer 8

so what two numbers are factors of two that add to get three?

Answer 9

2 and 1?

Answer 10

Yes!
So your new equation on top is \[(x+2)(x+1)\].
Now, let's do the bottom, it's a little bit harder.
\[(x ^{2}-2x-3)\]

Answer 11

So on this equation you must find the factors of NEGATIVE three, that add to get NEGATIVE 2. That means that one of the answers is going to be negative and one is going to be positive

Answer 12

\[(x-?)\times(x+?)=x ^{2}-2x-3\]

Answer 13

let me know if you get stuck

Answer 14

yeah im stuck

Answer 15

okay. So on this one it's going to be
\[(x-3)(x+1)\]
because -3 times +1 equals -3 and -3 plus +1 equals -2

Answer 16

so now you have the top and bottom of the first equation but you need to solve for the next equation. Overall this is what we have so far.
\[\frac{ (x+2)(x+1) }{ (x-3)(x+1) }\times \frac{ (x+?)(x+?) }{ (x+2) }\]

Answer 17

how do i find the other 2 numbers

Answer 18

@ab60093

Answer 19

i got that as my answer

Answer 20

but what are the restrictions

Answer 21

okay. lost my internet. im back now

Answer 22

its fine

Answer 23

ok well once you simplify everything you get the equation
\[\frac{ (x+2)(x+1) }{ (x-3)(x+1)}\times \frac{ (x+3)(x+1) }{ (x+2) }\]

Answer 24

And if you need to actually solve the problem you'd simplify and go from there but since you only need the restrictions we'll just do that

Answer 25

you have to set everything on the bottom equal to zero and solve for x. so...
\[(x-3)=0\]
\[(x+1)=0\]
\[(x+2)=0\]

Answer 26

x=3
x=-1
x=-2

Answer 27

well...those are your restrictions.
\[x \neq3, -1, -2\]

Answer 28

the problem is, idk why your answer choices have both 3 and -2 in there because that is the correct answer I even checked myself and graphed it

Answer 29

thats wierd what should i do then just pick one?

Answer 30

hold on

Answer 31

if you simplify out the equation it is equal to

therefore that would lead me to believe..
\[(x-3)=0\]
\[x \neq3\]
I would put answer choice 2.

Answer 32

okay thank you so much!! can i ask you one more question?

Answer 33

\[\frac{ (x+1)(x+3) }{ (x-3) }\]

Answer 34

for some reason that graph didn't go through...and sure

Answer 35

okay...do you know how to factor this one?

Answer 36

can you walk me through it again?

Answer 37

sure. So the top on has a 3 in front of the x meaning you should factor that out to make your life easier.so our equation should look something like this...
\[3(x+?)(x+?)\]
when you take out that 3 you must divide everything in that equation by 3 making the new equation
\[3(x ^{2}+4x+3)=3(x+3)(x+1)\]
that's your top equation

Answer 38

the bottom doesn't have a middle factor meaning that it is going to be the same number multiplied but one's positive and one's negative. Since it has to be the same number the square root of 4 is 2, so now on the bottom we have
\[(x+2)(x-2)=x ^{2}-4\]

Answer 39

so now everything has been factored. if you really wanted to you could turn
\[4x+4\rightarrow4(x+1)\] but you don't have to

Answer 40

so, to make this problem easier we're going to turn it into a multiplication problem. So do you know how to do that?

Answer 41

would you just multiply 4x+4 and 4(x+1)?

Answer 42

ummm. no. when you want to divide something that is being multiplied you take the reciprocal of one of the factors. For this problem we're turning
\[\frac{ 3(x+1)(x+3) }{ (x-2)(x+2) }\div \frac{ 4(x+1) }{ (x+2) }\]
into...
\[\frac{ 3(x+1)(x+3) }{ (x-2)(x+2) }\times \frac{ (x+2) }{ 4(x+1) }\]

Answer 43

you can also think of the
\[4(x+1)\] as \[4x+4\]

Answer 44

So now simplify it. cancel out the x=2 on the top and bottom and cancel out the 4(x+1) with the 3(x+3).
Your new equation is:
\[\frac{ (x+3) }{ (x-2) }\]
Now, do what we did earlier and put the bottom equal to zero and solve for x

Answer 45

so, what's the answer?

Answer 46

3?

Answer 47

no, remember you must set only the bottom equal to zero. so
\[(x-2)=0\] solve for x

Answer 48

so 2

Answer 49

yes. so your restriction is\[x \neq2\]

Answer 50

thank you again!

Answer 51

you're welcome :)

Answer 52

any more questions?

Answer 53

no thats it thanks so much