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mayankdevnani

  • 2 years ago

The remainder R obtained by dividing x^100 by x^2-3x+2 is a polynomial of degree less than 2.

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  1. Shadowys
    • 2 years ago
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    can't see the extreme right of the drawing though.

  2. mayankdevnani
    • 2 years ago
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    k

  3. sirm3d
    • 2 years ago
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    can you list the choices?

  4. mayankdevnani
    • 2 years ago
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    sure

  5. mayankdevnani
    • 2 years ago
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    |dw:1354621276560:dw|

  6. sirm3d
    • 2 years ago
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    |dw:1354621547979:dw|

  7. Shadowys
    • 2 years ago
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    @sirm3d i've got b though.

  8. sirm3d
    • 2 years ago
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    yes. it is b.

  9. Shadowys
    • 2 years ago
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    @mayankdevnani are you familiar with the remainder theorem?

  10. mayankdevnani
    • 2 years ago
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    @Shadowys and @sirm3d how is it b?

  11. mayankdevnani
    • 2 years ago
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    @sirm3d no, it is given that it is + sign

  12. Shadowys
    • 2 years ago
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    if you're familiar with the remainder theorem, go over here first: http://in.answers.yahoo.com/question/index?qid=20100116073124AA4yjsO

  13. mayankdevnani
    • 2 years ago
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    i know it

  14. Shadowys
    • 2 years ago
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    if you have seen the link, it is the same with this question. let \(f(x)=x^{100}\) since \(x^2-3x+2=(x-1)(x-2)\), we first use the RT for \(f(1)=1\) so, \(f(x)=(x-1)Q(x) +1\) you follow so far?

  15. mayankdevnani
    • 2 years ago
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    i don't understand!!!

  16. Shadowys
    • 2 years ago
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    which part of it? all of it? if yes then i'll adjust a bit. you've seen the link?

  17. mayankdevnani
    • 2 years ago
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    yes

  18. mayankdevnani
    • 2 years ago
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    i don't understand!!plz explain me step by step

  19. Shadowys
    • 2 years ago
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    let \(f(x) =x^{100}\) we want to find the remainder after dividing f(x) by \(x^2−3x+2\). using the remainder theorem, \(f(x^2−3x+2)=(x^2−3x+2)Q(x)+R\), right?

  20. mayankdevnani
    • 2 years ago
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    right

  21. Shadowys
    • 2 years ago
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    no, it's wrong. lol sorry about that \(f(x)=(x^2−3x+2)Q(x)+R\) so, \(f(x)=(x-1)(x-2)Q(x)+R\) now, notice that, if we sub x=1, the whole terms vanishes and we get f(1)=1,right?

  22. mayankdevnani
    • 2 years ago
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    i think it got -1

  23. mayankdevnani
    • 2 years ago
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    or 0

  24. mayankdevnani
    • 2 years ago
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    |dw:1354628302804:dw|

  25. Shadowys
    • 2 years ago
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    |dw:1354628370412:dw|

  26. Shadowys
    • 2 years ago
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    yup, the whole thing of Q(x) is just gone when x=1

  27. Shadowys
    • 2 years ago
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    and, by the remainder theorem, since f(1)=1, there must be a P(x) where \(f(x)=(x-1)P(x)+1\)

  28. Shadowys
    • 2 years ago
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    you follow?

  29. mayankdevnani
    • 2 years ago
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    yaa

  30. Shadowys
    • 2 years ago
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    notice that: \(f(x)=(x−1)P(x)+1\) and \(f(x)=(x−1)(x−2)Q(x)+R\) looks very similar. So, all we have to do is find a way to relate P(x) and Q(x). now, how do we do that? we let x=2 for the first eq. \(f(2)=(2−1)P(2)+1=2^{100}\) \(P(2)=2^{100}-1\) right?

  31. mayankdevnani
    • 2 years ago
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    right

  32. Shadowys
    • 2 years ago
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    now, since \(P(2)=2^{100 }-1\), by the remainder theorem again, there must exist a D(x) such that \(P(x)=(x-2)D(x)+2^{100 }-1\) !!! now, we substitute this into \(f(x)=(x−1)P(x)+1\), getting \(f(x)=(x−1)[(x-2)D(x)+2^{100 }-1]+1\) simplifying and rearranging, \(f(x)=(x−1)(x-2)D(x)+(x-1)(2^{100 }-1)+1\)

  33. Shadowys
    • 2 years ago
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    @sirm3d can you see where i went wrong here? i originally used x=2 first but x=1 should work too. @mayankdevnani start with x=2 and you'll get b. meanwhile sorry, lol x=1 didn't work out.

  34. mayankdevnani
    • 2 years ago
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    thnx... @Shadowys

  35. Shadowys
    • 2 years ago
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    you're welcome :) though i messed up a lil

  36. sirm3d
    • 2 years ago
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    i see nothing wrong here. \[ (x-1)(2^{100}-1)+1=2^{100}(x-1)-(x-1)+1=2^{100}(x-1)-(x-2)\]

  37. Shadowys
    • 2 years ago
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    oh. I forgot about that -1 lol i guess I shouldn't do maths during bedtime lol

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