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mayankdevnani
Group Title
The remainder R obtained by dividing x^100 by x^23x+2 is a polynomial of degree less than 2.
 one year ago
 one year ago
mayankdevnani Group Title
The remainder R obtained by dividing x^100 by x^23x+2 is a polynomial of degree less than 2.
 one year ago
 one year ago

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Shadowys Group TitleBest ResponseYou've already chosen the best response.3
can't see the extreme right of the drawing though.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
can you list the choices?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
sure
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
dw:1354621276560:dw
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
dw:1354621547979:dw
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
@sirm3d i've got b though.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
yes. it is b.
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
@mayankdevnani are you familiar with the remainder theorem?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@Shadowys and @sirm3d how is it b?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@sirm3d no, it is given that it is + sign
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
if you're familiar with the remainder theorem, go over here first: http://in.answers.yahoo.com/question/index?qid=20100116073124AA4yjsO
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
i know it
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
if you have seen the link, it is the same with this question. let \(f(x)=x^{100}\) since \(x^23x+2=(x1)(x2)\), we first use the RT for \(f(1)=1\) so, \(f(x)=(x1)Q(x) +1\) you follow so far?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
i don't understand!!!
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
which part of it? all of it? if yes then i'll adjust a bit. you've seen the link?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
i don't understand!!plz explain me step by step
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
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?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
right
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
no, it's wrong. lol sorry about that \(f(x)=(x^2−3x+2)Q(x)+R\) so, \(f(x)=(x1)(x2)Q(x)+R\) now, notice that, if we sub x=1, the whole terms vanishes and we get f(1)=1,right?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
i think it got 1
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
or 0
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
dw:1354628302804:dw
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
dw:1354628370412:dw
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
yup, the whole thing of Q(x) is just gone when x=1
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
and, by the remainder theorem, since f(1)=1, there must be a P(x) where \(f(x)=(x1)P(x)+1\)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
you follow?
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
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?
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
right
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
now, since \(P(2)=2^{100 }1\), by the remainder theorem again, there must exist a D(x) such that \(P(x)=(x2)D(x)+2^{100 }1\) !!! now, we substitute this into \(f(x)=(x−1)P(x)+1\), getting \(f(x)=(x−1)[(x2)D(x)+2^{100 }1]+1\) simplifying and rearranging, \(f(x)=(x−1)(x2)D(x)+(x1)(2^{100 }1)+1\)
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
@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.
 one year ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
thnx... @Shadowys
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
you're welcome :) though i messed up a lil
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
i see nothing wrong here. \[ (x1)(2^{100}1)+1=2^{100}(x1)(x1)+1=2^{100}(x1)(x2)\]
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
oh. I forgot about that 1 lol i guess I shouldn't do maths during bedtime lol
 one year ago
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