Question

I WILL FAN/MEDAL.
Divide. (x^2-16/x-1)
a. x-4/x-1
b.x+4/x-1
c.(x+4)(x-3)/x-1
d. x-4/x+1

Answer 1

Quick question...for C...is that supposed to be

\[\large \frac{(x + 4)(x - 4)}{x - 1}\]

Answer 2

No! I thought so too, but maybe it's just a typo in the study guide?!

Answer 3

Must be because (although this is giving away an answer) that would be it :)

Answer 4

Haha well thank you for helping! Do you think you could help me out on another one?

Answer 5

Lol even though you already knew :P

And yeah sure

Answer 6

True, but it's good to have some reassurance!
Divide.

Answer 7

I have to type the answer choices they're gonna take a while haha I suck at this

Answer 8

Lol you're fine :)

So we have

\[\huge \frac{\frac{x^2 + 2x + 1}{x - 2}}{\frac{x^2 - 1}{x^2 - 4}}\]

right?

Answer 9

and yes! I'm totally stuck on this one. I have no clue how to even start it >.<

Answer 10

Well...we can do 1 of 2 things first

We can turn everything into factors...

Or

we can start right off the bat turning this into multiplication instead of division ...your pick :D

Answer 11

Multiplication PLEASE! I suck in division :P

Answer 12

Lol okay...well first remember what happens when you divide by a fraction

\[\huge \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}\]

How? Well...to make it sound easy you flip the bottom fraction...and you multiply

Here, we have

\[\huge \frac{\frac{x^2 + 2x + 1}{x - 2}}{\frac{x^2 - 1}{x^2 - 4}}\]

If we flip the bottom fraction....and we multiply everything by it, we get

\[\large \frac{(x^2 + 2x + 1)(x^2 - 4)}{(x - 2)(x^2 - 1)}\]

Let me know if you're okay with that...and we can proceed :)

Answer 13

I understand! But is this for every division problem or just the ones with exponents?!

Answer 14

Whenever you are dividing by a fraction you would do this process...it helps keep everything organized lol

But you understand it, so you'll be good :D

\[\large \frac{(x^2 + 2x + 1)(x^2 - 4)}{(x - 2)(x^2 - 1)}\]

What would we do next?

Answer 15

Divide exponents first right?

Answer 16

Or would it be parenthesis first?

Answer 17

Oh neither...that would be making it much more complicated that it needs to be

Hint* Factor everything that can be factored :)

Answer 18

Oh wow that's what they taught in school -.-
And would that be the 2 and 6?

Answer 19

The 2 and the 6? Not quite sure what you mean?

Just go along with the process like before in that last question where you knew

\[\large x^2- 16 = (x+4)(x-4)\]

Apply that to this question :)

Answer 20

Okay so when you say factor do you mean everything that's the same number? Or everything that can go into a number equally? I'm sorry for all the questions I only have one semester of this class and I haven't been doing very good. I'm supposed to be doing this study guide but I can't figure out these division problems to save my life -.-

Answer 21

Not a problem :D that's why we're here lol

Okay

Going back to your last question....how did you know that

\[\large x^2 - 16 = (x+4)(x-4)\]?

Answer 22

I know the x has a power of two because there's two x's present in this problem, and 4 mult. by 4 gives you 16. since the negative sign is left over it gets put in with the x^2-16 right?

Answer 23

Well, kind of, yes and no :)

So starting from the original quadratic

\[\large x^2 + 0x - 16\]

Notice...I have written in '0x' just as a placeholder and for explanation...it is STILL just x^2 - 16 right?

Answer 24

So from here, we need to find 2 numbers

Those 2 numbers, when multiplied together will make -16
But at the same time...they will add together and make 0

Answer 25

True, so is that why that makes it negative? the -16?

Answer 26

Now we know that -4 times 4 = 16

And also -4 + 4 = 0...so those are our 2 numbers that we will use

We can now write the "factors" of \(\large x^2 - 16\)

So

\[\large x^2 - 16 = (x + \text{one number we just found} )(x + \text{other number we just found})\]

Meaning

\[\large x^2 - 16 = (x + 4)(x - 4)\]

Answer 27

Sorry typo up there

*we know that -4 times 4 = -16 !! forgot the negative sign!!

Answer 28

I hope that made sense?

Answer 29

Yes it does!!! thank you for further explaining it to me! So applying what you just said to the next problem would the factors be x^2?

Answer 30

Perfect...

Okay...

So lets focus on....you have a (x^2 - 1) up there

How can we apply this approach to that?

\[\large x^2 + 0x - 1\]

Answer 31

so that would be x^2-1?
or would the +x make it x^3?

Answer 32

Nope...remember that 0x is just there as a placeholder...nothing gets combined :)

From the equation

\[\large x^2 + 0x - 1\]

We want 2 numbers...that when multiplied make -1
but at the same time, when added make 0

What are the 2 numbers?

Answer 33

I don't know if my calculator is wrong or if I'm doing it wrong.
I did -1+0 and got -1, I did x2+x and got 1 -____-

Answer 34

I'm thinking its 0x+x2 because there's no numbers there so if added it would still be 0?

Answer 35

Well this isn't really something that can be done in a calculator unless you have the cool one XD

But here we just think

Okay, what 2 numbers multiply to make -1? hmm, well I know -1 and 1 multiply to make -1

But do they add to make 0? well -1 + 1 = 0

So yes...those are my 2 numbers....is that making sense?

Answer 36

Okay! I thought we had to use two different numbers I didn't think we could use the same one >.< lol, yes it does make sense now!

Answer 37

Oh no no, they can be whatever numbers fit the situation :)

Okay so

\[\large x^2 - 1 = (x + 1)(x - 1)\]

Good? make sure because we move to the next one after this :P

Answer 38

yes that makes total sense! I tried solving the next problem but I'm not sure if I did it right. Do you want to check it and tell me where I went wrong?( If I did.)

Answer 39

Of course :)

Answer 40

okay so I broke the problem up into sections and tried doing it that way.