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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    This was the result I got when multiplying two fractions. I know my multiplication was correct.

  2. anonymous
    • 5 years ago
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    No.

  3. anonymous
    • 5 years ago
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    Why?

  4. anonymous
    • 5 years ago
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    I'm guessing you'd have to solve the top and the bottom separately for it to be simplified.

  5. anonymous
    • 5 years ago
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    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?

  6. anonymous
    • 5 years ago
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    Or just the -36/48 part...

  7. anonymous
    • 5 years ago
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    I believe its just the -36/48 part.

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

  9. anonymous
    • 5 years ago
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    ?

  10. anonymous
    • 5 years ago
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    Lost me their, man. I'm only in 8th grade.

  11. anonymous
    • 5 years ago
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    Eh... Does that look completely simplified to you then?

  12. anonymous
    • 5 years ago
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    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...

  13. anonymous
    • 5 years ago
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    yo no entiendo las cosas que este chico esta preguntando me...

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

  15. anonymous
    • 5 years ago
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    K.

  16. anonymous
    • 5 years ago
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    Darn it. I messed it up. I can't solve that right now. I'm studying at the same time I do this.

  17. anonymous
    • 5 years ago
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    Sorry man. You're on your own until someone with the proper knowledge can help you. :/

  18. anonymous
    • 5 years ago
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    Si no sabes cómo hacerlo, ¿por qué estás tratando de ayudar?

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

  20. anonymous
    • 5 years ago
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    yes

  21. anonymous
    • 5 years ago
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    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.

  22. anonymous
    • 5 years ago
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    It does not look difficult at all. Just a minute.

  23. anonymous
    • 5 years ago
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    such as -2x^3/6x^3, would you simplify that to x^3/3x^3?

  24. anonymous
    • 5 years ago
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    I forgot the negative in that answer, sorry

  25. anonymous
    • 5 years ago
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    Are you sure it's division, not multiplication?

  26. anonymous
    • 5 years ago
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    Absolutely positive

  27. anonymous
    • 5 years ago
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    I flipped the fraction on the right side to turn the division sign into a multiplication sign.

  28. anonymous
    • 5 years ago
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    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.

  29. anonymous
    • 5 years ago
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    Ahhh, I got on here yesterday to ask about factoring it and someone told me not to do that.

  30. anonymous
    • 5 years ago
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    But you know this type of question usually involve some cancellation, which would be the case if it was a multiplication.

  31. anonymous
    • 5 years ago
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    But I get why you did it. It IS more simple in factored form.

  32. anonymous
    • 5 years ago
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    From what I remember of my first attempt at factoring it, all of the factored trinomials did not cancel.

  33. anonymous
    • 5 years ago
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    I have the work around here, just one secon.

  34. anonymous
    • 5 years ago
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    Yeah, nothing can be cancelled. Is this a question from your teacher or from the book?

  35. anonymous
    • 5 years ago
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    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.

  36. anonymous
    • 5 years ago
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    I see. Good luck. I'll be around of you have any questions that I have answers for.

  37. anonymous
    • 5 years ago
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    But in that form you gave, couldn't you factor it further?

  38. anonymous
    • 5 years ago
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    Nope.

  39. anonymous
    • 5 years ago
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    but x^2-2x+4 would be (x-2)(x-2) wouldn't it?

  40. anonymous
    • 5 years ago
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    Wait no, it won't work that way.

  41. anonymous
    • 5 years ago
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    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.

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

  43. anonymous
    • 5 years ago
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    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?

  44. anonymous
    • 5 years ago
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    But what I wrote before is more simplified than this, I would say.

  45. anonymous
    • 5 years ago
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    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

  46. anonymous
    • 5 years ago
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    You're welcome, but you can't cancel them. :)

  47. anonymous
    • 5 years ago
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    if there was multiplication in the center you could though, right?

  48. anonymous
    • 5 years ago
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    Yeah.

  49. anonymous
    • 5 years ago
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    You could cancel many things actually.

  50. anonymous
    • 5 years ago
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    Like this: \[(x^2-2x+4) * (1)\] \[(1) (x+2)(x^2-2x+4)\]

  51. anonymous
    • 5 years ago
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    eh, everything after the 1 is on the right side of that problem. Cant figure out how to like up the equation things

  52. anonymous
    • 5 years ago
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    It would seem like, in that scenario, you would have 1/1 * 1/(x+2)

  53. anonymous
    • 5 years ago
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    I don't get it :)

  54. anonymous
    • 5 years ago
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    Sorry, try it this way: 1 5 - * - 5 1

  55. anonymous
    • 5 years ago
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    Yeah this is 1.

  56. anonymous
    • 5 years ago
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    I am applying that principle to the fractions. You helped me figure it out though :D Thank you again!

  57. anonymous
    • 5 years ago
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    :)

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