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

- anonymous

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- anonymous

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

- anonymous

No.

- anonymous

Why?

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## More answers

- anonymous

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

- anonymous

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?

- anonymous

Or just the -36/48 part...

- anonymous

I believe its just the -36/48 part.

- anonymous

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

- anonymous

?

- anonymous

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

- anonymous

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

- anonymous

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...

- anonymous

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

- anonymous

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

- anonymous

K.

- anonymous

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

- anonymous

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

- anonymous

Si no sabes cÃ³mo hacerlo, Â¿por quÃ© estÃ¡s tratando de ayudar?

- anonymous

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

- anonymous

yes

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

I forgot the negative in that answer, sorry

- anonymous

Are you sure it's division, not multiplication?

- anonymous

Absolutely positive

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I have the work around here, just one secon.

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

Nope.

- anonymous

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

- anonymous

Wait no, it won't work that way.

- anonymous

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.

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Yeah.

- anonymous

You could cancel many things actually.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I don't get it :)

- anonymous

Sorry, try it this way:
1 5
- * -
5 1

- anonymous

Yeah this is 1.

- anonymous

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

- anonymous

:)

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