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

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

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

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

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

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

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

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

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

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
i know it
 2 years 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?
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
i don't understand!!!
 2 years 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?
 2 years ago

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

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
right
 2 years 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?
 2 years ago

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

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

Shadowys Group TitleBest ResponseYou've already chosen the best response.3
dw:1354628370412:dw
 2 years 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
 2 years 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 \(f(x)=(x1)P(x)+1\)
 2 years ago

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

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
right
 2 years 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, \(f(x)=(x−1)(x2)D(x)+(x1)(2^{100 }1)+1\)
 2 years 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.
 2 years ago

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

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

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