anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
This was the result I got when multiplying two fractions. I know my multiplication was correct.
anonymous
  • anonymous
No.
anonymous
  • anonymous
Why?

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anonymous
  • anonymous
I'm guessing you'd have to solve the top and the bottom separately for it to be simplified.
anonymous
  • 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
  • anonymous
Or just the -36/48 part...
anonymous
  • anonymous
I believe its just the -36/48 part.
anonymous
  • 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
?
anonymous
  • anonymous
Lost me their, man. I'm only in 8th grade.
anonymous
  • anonymous
Eh... Does that look completely simplified to you then?
anonymous
  • 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
  • anonymous
yo no entiendo las cosas que este chico esta preguntando me...
anonymous
  • 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
  • anonymous
K.
anonymous
  • 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
  • anonymous
Sorry man. You're on your own until someone with the proper knowledge can help you. :/
anonymous
  • anonymous
Si no sabes cómo hacerlo, ¿por qué estás tratando de ayudar?
anonymous
  • anonymous
Is this the original expression \[{x^2-2x+4 \over x^2-5x+6}\div {x^2-9 \over x^3+8}?\]
anonymous
  • anonymous
yes
anonymous
  • 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
  • anonymous
It does not look difficult at all. Just a minute.
anonymous
  • anonymous
such as -2x^3/6x^3, would you simplify that to x^3/3x^3?
anonymous
  • anonymous
I forgot the negative in that answer, sorry
anonymous
  • anonymous
Are you sure it's division, not multiplication?
anonymous
  • anonymous
Absolutely positive
anonymous
  • anonymous
I flipped the fraction on the right side to turn the division sign into a multiplication sign.
anonymous
  • 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
  • anonymous
Ahhh, I got on here yesterday to ask about factoring it and someone told me not to do that.
anonymous
  • anonymous
But you know this type of question usually involve some cancellation, which would be the case if it was a multiplication.
anonymous
  • anonymous
But I get why you did it. It IS more simple in factored form.
anonymous
  • anonymous
From what I remember of my first attempt at factoring it, all of the factored trinomials did not cancel.
anonymous
  • anonymous
I have the work around here, just one secon.
anonymous
  • anonymous
Yeah, nothing can be cancelled. Is this a question from your teacher or from the book?
anonymous
  • 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
  • anonymous
I see. Good luck. I'll be around of you have any questions that I have answers for.
anonymous
  • anonymous
But in that form you gave, couldn't you factor it further?
anonymous
  • anonymous
Nope.
anonymous
  • anonymous
but x^2-2x+4 would be (x-2)(x-2) wouldn't it?
anonymous
  • anonymous
Wait no, it won't work that way.
anonymous
  • 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
  • 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
  • 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
  • anonymous
But what I wrote before is more simplified than this, I would say.
anonymous
  • 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
  • anonymous
You're welcome, but you can't cancel them. :)
anonymous
  • anonymous
if there was multiplication in the center you could though, right?
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
You could cancel many things actually.
anonymous
  • anonymous
Like this: \[(x^2-2x+4) * (1)\] \[(1) (x+2)(x^2-2x+4)\]
anonymous
  • 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
  • anonymous
It would seem like, in that scenario, you would have 1/1 * 1/(x+2)
anonymous
  • anonymous
I don't get it :)
anonymous
  • anonymous
Sorry, try it this way: 1 5 - * - 5 1
anonymous
  • anonymous
Yeah this is 1.
anonymous
  • anonymous
I am applying that principle to the fractions. You helped me figure it out though :D Thank you again!
anonymous
  • anonymous
:)

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