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Thread: Help!

  1. #11
    Elite Member stapel's Avatar
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    Quote Originally Posted by BabyBlue
    People,
    I am still here
    Okay.... Did you have a question...?

    You've been provided with half of the worked solution, and tutors have attempted to figure out how to answer your other (somewhat befuddling) posts. If you are needing help with other details of this process (like how to combine "like" terms, or how to do arithmetic with negatives, or how to multiply polynomials, etc), then please reply with specifics.

    Thank you!

    Eliz.

  2. #12
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    Quote Originally Posted by stapel
    Quite possibly not. Modern progressive (new-new) educationist thought ...
    Well, there I am, out of the loop again.

    In any case, I wanted to back off my position just a bit. The numerical long division uses only integers and uses the entire structure of the divisor in each step. The polynomial version is severely generalized, caring only about the first element at each phase and allowing noninteger and negative results. I think a numerical long division would have a problem with the answer 1()(-3)4.

  3. #13
    Elite Member stapel's Avatar
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    Quote Originally Posted by tkhunny
    The numerical long division uses only integers and uses the entire structure of the divisor in each step.
    Roughly, yes. But if you look closely at the raw process, you'll note that the way you often find your first guess (in "A B"), as to how many times B will go into A, is to look at the first few (leading) digits of A and the first few digits, or rounded value, of B.

    For instance, in dividing 295349 by 47, you wouldn't necessarily initially consider all of 295349 or the exact value 47. Instead, you might note that 47 is bigger than 29, so you won't be putting "1" on top of the long-division sign. Adding in another digit, you have 295 and 47. This is, roughly, 300 and 50, so try 6: (47)(6) = 282. This "fits" inside 295, so put "6" on top, put "282" underneath, and subtract.

    Lather. Rinse. Repeat.

    Eliz.

  4. #14
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    Quote Originally Posted by stapel
    Lather. Rinse. Repeat.

    Eliz.
    That's wonderful, Eliz!

  5. #15
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    I am trying to be patient, but I do not like being laughed at. I am trying to learn.
    I'll submit the answer soon.
    trying hard

  6. #16
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    doesn't look like anybody has laughed or made fun of you in this thread ... in fact, not a semblance of mockery/derision can be found in the posts above ...

  7. #17
    Elite Member stapel's Avatar
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    Quote Originally Posted by Amongst the various partial solutions and explanations, one of the tutors
    ...I am glad to see that you are trying to be a bit more down-to-earth about [this].
    Quote Originally Posted by BabyBlue
    I am trying to be patient, but I do not like being laughed at.
    It is to be regretted if you feel that the step-by-step instructions, the explanations, and the encouragement you've been offered are somehow meant to mock you. But I'm afraid there is little we can do to fix that particular situation. Sorry!

    My best wishes to you in your studies!

    Eliz.

  8. #18
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    Sorry if I took offense to easily :P
    trying hard

  9. #19
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    my answer

    The answer is:
    a^7-a^6+2a^5-2a^4+3a^3-3a^2-3a
    fraction -1 over a+1
    trying hard

  10. #20
    Elite Member stapel's Avatar
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    Quote Originally Posted by BabyBlue
    The answer is:
    a^7-a^6+2a^5-2a^4+3a^3-3a^2-3a
    fraction -1 over a+1
    Do you mean that this was the solution that was provided to you? Or that this is what you've come up with, and you're requesting correction or confirmation?

    Do you mean the following?

    . . . . .a^7 - a^6 + 2a^5 - 2a^4 + 3a^3 - 3a^2 - 3a - 1/(a + 1)

    Or do you mean something else?

    Thank you!

    Eliz.

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