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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Jan 2008 18:24:14 IST
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Could anyone please tell me how to solve problems of this type? If f(xn) is divisible by (x-1), it is also divisible by? (I know this looks really a silly question but i dunno how to get started!)
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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17 Jan 2008 18:57:14 IST
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One possible answer: (x-z) where z is a solution of zn = 1; z is Complex.
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19 Jan 2008 17:56:24 IST
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Hey cmon sum1 reply!!
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20 Jan 2008 09:26:29 IST
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anything wrong with my answer?
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20 Jan 2008 11:44:02 IST
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Ok i found the answer...its xn. But i still don;t know the detailed solution
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5 Feb 2008 00:28:46 IST
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well saw this question only now. Are you sure thats the entire question? If n=1 and f(x) = x - 1, the result doesnt hold...
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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5 Feb 2008 08:13:29 IST
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I'll ask once again is anything wrong with answer that f(x) is also divisible by (x-z) where z is a solution to x n=1 and z  C
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5 Feb 2008 12:04:29 IST
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Heres the solution... f(1n)=0 by factor theorem putting xn=y f(xn)=f(y) with f(1)=0 => y-1 is a factor of f(y) => (xn-1)is a factor of f(xn)
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
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5 Feb 2008 12:12:15 IST
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plz dont rate...copied the soln from book
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- Bill Gates |
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5 Feb 2008 13:16:26 IST
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I still maintain that my answer is right. any naysayers? The qn only asks us to find any factor of f(xn) not particularly to prove that (xn-1) is a factor. You can easily see that (xn-1) is a factor automatically implies that (x-z) where z is a solution to zn=1 becomes a factor. @Saurabh: Let f(x) = anxn+an-1xn-1+...+a0 Then f(xn) = an(xn)n + an-1xn-1+..+a0. f(1) = an+an-1+...+a0 = 0 as x-1 is a solution Now you consider any solution to zn=1 and find f(xn) where x = z you get f(zn) = f(1) = 0 f(xn) = F(x) where F is a polynomial of degree n2 (of which an is the coefficient). and F(z) = 0. Hence (x-z) is a factor of F(x) and hence of f(xn).
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5 Feb 2008 13:37:02 IST
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Akhil pls tell me how did u agree with : (y-1) is a factor of f(y) ? ok tell me to which chapter the question posted by you belongs ? & from which book ?
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EVEN SKY IS NOT THE LIMIT...FLY HIGHER IF YOU CAN .! |
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5 Feb 2008 13:41:30 IST
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Its from Brilliants YG file (old one)-algebra
i wasnt too sure of their solution either so i posted it here
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" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
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5 Feb 2008 13:47:48 IST
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ok
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EVEN SKY IS NOT THE LIMIT...FLY HIGHER IF YOU CAN .! |
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5 Feb 2008 13:52:54 IST
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nice solution akhil.. and of course, hsbhatt's is also correct. but I still maintain the above is valid only for n  1. |
Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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