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

- mayankdevnani

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

- katieb

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- anonymous

can't see the extreme right of the drawing though.

- mayankdevnani

k

- sirm3d

can you list the choices?

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## More answers

- mayankdevnani

sure

- mayankdevnani

|dw:1354621276560:dw|

- sirm3d

|dw:1354621547979:dw|

- anonymous

@sirm3d i've got b though.

- sirm3d

yes. it is b.

- anonymous

@mayankdevnani are you familiar with the remainder theorem?

- mayankdevnani

@Shadowys and @sirm3d how is it b?

- mayankdevnani

@sirm3d no, it is given that it is + sign

- anonymous

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

- mayankdevnani

i know it

- 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

i don't understand!!!

- anonymous

which part of it? all of it? if yes then i'll adjust a bit.
you've seen the link?

- mayankdevnani

yes

- mayankdevnani

i don't understand!!plz explain me step by step

- 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

right

- 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

i think it got -1

- mayankdevnani

or 0

- mayankdevnani

|dw:1354628302804:dw|

- anonymous

|dw:1354628370412:dw|

- anonymous

yup, the whole thing of Q(x) is just gone when x=1

- 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

you follow?

- mayankdevnani

yaa

- 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

right

- 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

@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

thnx... @Shadowys

- anonymous

you're welcome :) though i messed up a lil

- 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

oh. I forgot about that -1 lol i guess I shouldn't do maths during bedtime lol

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