mayankdevnani
  • mayankdevnani
The remainder R obtained by dividing x^100 by x^2-3x+2 is a polynomial of degree less than 2.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
can't see the extreme right of the drawing though.
mayankdevnani
  • mayankdevnani
k
sirm3d
  • sirm3d
can you list the choices?

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More answers

mayankdevnani
  • mayankdevnani
sure
mayankdevnani
  • mayankdevnani
|dw:1354621276560:dw|
sirm3d
  • sirm3d
|dw:1354621547979:dw|
anonymous
  • anonymous
@sirm3d i've got b though.
sirm3d
  • sirm3d
yes. it is b.
anonymous
  • anonymous
@mayankdevnani are you familiar with the remainder theorem?
mayankdevnani
  • mayankdevnani
@Shadowys and @sirm3d how is it b?
mayankdevnani
  • mayankdevnani
@sirm3d no, it is given that it is + sign
anonymous
  • anonymous
if you're familiar with the remainder theorem, go over here first: http://in.answers.yahoo.com/question/index?qid=20100116073124AA4yjsO
mayankdevnani
  • mayankdevnani
i know it
anonymous
  • anonymous
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?
mayankdevnani
  • mayankdevnani
i don't understand!!!
anonymous
  • anonymous
which part of it? all of it? if yes then i'll adjust a bit. you've seen the link?
mayankdevnani
  • mayankdevnani
yes
mayankdevnani
  • mayankdevnani
i don't understand!!plz explain me step by step
anonymous
  • anonymous
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?
mayankdevnani
  • mayankdevnani
right
anonymous
  • anonymous
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?
mayankdevnani
  • mayankdevnani
i think it got -1
mayankdevnani
  • mayankdevnani
or 0
mayankdevnani
  • mayankdevnani
|dw:1354628302804:dw|
anonymous
  • anonymous
|dw:1354628370412:dw|
anonymous
  • anonymous
yup, the whole thing of Q(x) is just gone when x=1
anonymous
  • anonymous
and, by the remainder theorem, since f(1)=1, there must be a P(x) where \(f(x)=(x-1)P(x)+1\)
anonymous
  • anonymous
you follow?
mayankdevnani
  • mayankdevnani
yaa
anonymous
  • anonymous
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?
mayankdevnani
  • mayankdevnani
right
anonymous
  • anonymous
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, \(f(x)=(x−1)(x-2)D(x)+(x-1)(2^{100 }-1)+1\)
anonymous
  • anonymous
@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.
mayankdevnani
  • mayankdevnani
thnx... @Shadowys
anonymous
  • anonymous
you're welcome :) though i messed up a lil
sirm3d
  • sirm3d
i see nothing wrong here. \[ (x-1)(2^{100}-1)+1=2^{100}(x-1)-(x-1)+1=2^{100}(x-1)-(x-2)\]
anonymous
  • anonymous
oh. I forgot about that -1 lol i guess I shouldn't do maths during bedtime lol

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