Question

Would the following fraction be considered to be fully simplified?

x^4-2x^3-5x^2-18x-36
-----------------------------
x^5-5x^4+6x^3+8x^2-40x+48

Answer 1

This was the result I got when multiplying two fractions. I know my multiplication was correct.

Answer 2

No.

Answer 3

Why?

Answer 4

I'm guessing you'd have to solve the top and the bottom separately for it to be simplified.

Answer 5

No, there is no solving. My instructions are to simplify completely. This is where I am in the problem. Does that mean I need to combine all of the x's and the whole numbers in front of them, and then subtract the smaller total from the larger total, and get an x with a single variable on whatever side of the equation that happens to be?

Answer 6

Or just the -36/48 part...

Answer 7

I believe its just the -36/48 part.

Answer 8

Okay yea, so then that would make the final answer:
\[x^4-2x^3-5x^2-18x-3 \]
------------------------
\[x^5-5x^4+6x^3+8x^2-40x+12\]

Answer 9

?

Answer 10

Lost me their, man. I'm only in 8th grade.

Answer 11

Eh... Does that look completely simplified to you then?

Answer 12

No. Doesn't look simplified. I would just try out all the options and see which one does look like its in its final "simplified" form...

Answer 13

yo no entiendo las cosas que este chico esta preguntando me...

Answer 14

Okay, then how about the original question... See what you can come up with...
\[(x^2-2x+4)/(x^2-5x+6)\div(x^2-9)/(x^3+8)\]

Answer 15

K.

Answer 16

Darn it. I messed it up. I can't solve that right now. I'm studying at the same time I do this.

Answer 17

Sorry man. You're on your own until someone with the proper knowledge can help you. :/

Answer 18

Si no sabes cómo hacerlo, ¿por qué estás tratando de ayudar?

Answer 19

Is this the original expression
\[{x^2-2x+4 \over x^2-5x+6}\div {x^2-9 \over x^3+8}?\]

Answer 20

yes

Answer 21

At the very top is what I got from long multiplication. I just need to know if that form is correctly simplified since the exponents on the variables aren't all the same, or if I have to simplify those variables with the same exponents if possible.

Answer 22

It does not look difficult at all. Just a minute.

Answer 23

such as -2x^3/6x^3, would you simplify that to x^3/3x^3?

Answer 24

I forgot the negative in that answer, sorry

Answer 25

Are you sure it's division, not multiplication?

Answer 26

Absolutely positive

Answer 27

I flipped the fraction on the right side to turn the division sign into a multiplication sign.

Answer 28

No problem.
\[={x^2-2x+4 \over (x-3)(x-2)}.{(x+2)(x^2-2x+4) \over (x-3)(x+3)}\]
\[{(x^2-2x+4)^2(x+2) \over (x-3)^2(x+3)(x-2)}\]
For me, this is the best simplification.

Answer 29

Ahhh, I got on here yesterday to ask about factoring it and someone told me not to do that.

Answer 30

But you know this type of question usually involve some cancellation, which would be the case if it was a multiplication.

Answer 31

But I get why you did it. It IS more simple in factored form.

Answer 32

From what I remember of my first attempt at factoring it, all of the factored trinomials did not cancel.

Answer 33

I have the work around here, just one secon.

Answer 34

Yeah, nothing can be cancelled. Is this a question from your teacher or from the book?

Answer 35

It is a study guide for my final exam for college algebra. There are no answers given, so I want to make sure I have the idea down.

Answer 36

I see. Good luck. I'll be around of you have any questions that I have answers for.

Answer 37

But in that form you gave, couldn't you factor it further?

Answer 38

Nope.

Answer 39

but x^2-2x+4 would be (x-2)(x-2) wouldn't it?

Answer 40

Wait no, it won't work that way.

Answer 41

But I get what you are saying. Factoring would be the simplest way to completely simplify a problem with two fractions and any kind of operation.

Answer 42

Well there is one thing you could do to the denominator. The denominator can be written as (x^2-9)(x-3)(x-2)

Answer 43

Actually, (x^2-2x+4)/1 would cross cancel with 1/(x+2)(x^2-2x+4) which is the result of x^3+8, right?

Answer 44

But what I wrote before is more simplified than this, I would say.

Answer 45

Yes, and I appreciate the help very much, I will let you move on to other things, but you helped me figure something out :D

Answer 46

You're welcome, but you can't cancel them. :)

Answer 47

if there was multiplication in the center you could though, right?

Answer 48

Yeah.

Answer 49

You could cancel many things actually.

Answer 50

Like this:

\[(x^2-2x+4) * (1)\]
\[(1) (x+2)(x^2-2x+4)\]

Answer 51

eh, everything after the 1 is on the right side of that problem. Cant figure out how to like up the equation things

Answer 52

It would seem like, in that scenario, you would have 1/1 * 1/(x+2)

Answer 53

I don't get it :)

Answer 54

Sorry, try it this way:

1 5
- * -
5 1

Answer 55

Yeah this is 1.

Answer 56

I am applying that principle to the fractions. You helped me figure it out though :D Thank you again!

Answer 57

:)