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.

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

2. mayankdevnani

k

3. sirm3d

can you list the choices?

4. mayankdevnani

sure

5. mayankdevnani

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6. sirm3d

|dw:1354621547979:dw|

@sirm3d i've got b though.

8. sirm3d

yes. it is b.

@mayankdevnani are you familiar with the remainder theorem?

10. mayankdevnani

@Shadowys and @sirm3d how is it b?

11. mayankdevnani

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

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

13. mayankdevnani

i know it

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

i don't understand!!!

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

17. mayankdevnani

yes

18. mayankdevnani

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

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

right

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

i think it got -1

23. mayankdevnani

or 0

24. mayankdevnani

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

and, by the remainder theorem, since f(1)=1, there must be a P(x) where \(f(x)=(x-1)P(x)+1\)

you follow?

29. mayankdevnani

yaa

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

right

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, \(f(x)=(x−1)(x-2)D(x)+(x-1)(2^{100 }-1)+1\)

@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

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

36. sirm3d

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