The reminder when polynomial 1+x^2 +x^4 +x^6 +....+x^(22) is divided by 1+x^ +x^2 +x^3 +.....+x^(11) is =?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 46959 by rahul 19 last updated on 03/Nov/18

$T h e r e m i n d e r w h e n p o l y n o m i a l$ $1 + x^{2} + x^{4} + x^{6} + . . . . + x^{22} i s d i v i d e d b y$ $1 + x^{} + x^{2} + x^{3} + . . . . . + x^{11} i s = ?$
Commented by rahul 19 last updated on 03/Nov/18

$I^{'} m g e t t i n g 2 w h i c h i s w r o n g!$ $w e c a n s e e t h e y b o t h f o r m s a G . P$ $\Rightarrow r e m a i n d e r w h e n \frac{x^{12} + 1}{x + 1} = 2$ $(c a n c e l l i n g c o m m o n t e r m s . . . .)$ $W h e r e a m i w r o n g ? ?$
Commented by maxmathsup by imad last updated on 03/Nov/18

$l e t p (x) = 1 + x^{2} + x^{4} + . . . + x^{22} \Rightarrow p (x) = 1 + x^{2} + {(x^{2})}^{2} + . . . {(x^{2})}^{11}$ $= \frac{1 - {(x^{2})}^{12}}{1 - x^{2}} i f x^{2} \neq 1 \Rightarrow p (x) = \frac{x^{24} - 1}{x^{2} - 1} l e t q (x) = 1 + x + x^{2} + . . . x^{11} \Rightarrow$ $q (x) = \frac{1 - x^{12}}{1 - x} i f x \neq 1 \Rightarrow \frac{p (x}{q (x)} = \frac{x^{24} - 1}{x^{2} - 1} \frac{x - 1}{x^{12} - 1} = \frac{x^{24} - 1}{(x + 1) (x^{12} - 1)}$ $= \frac{{(x^{12})}^{2} - 1}{(x + 1) (x^{12} - 1)} = \frac{x^{12} + 1}{x + 1}$ $x^{12} + 1 = 0 \Rightarrow x^{12} = e^{i (2 k + 1) π} \Rightarrow t h e r o o t s a r e x_{k} = e^{i \frac{(2 k + 1) π}{12}} k \in [[0, 11]] \Rightarrow$ $x^{12} + 1 = \prod_{k = 0}^{11} (x - e^{i \frac{(2 k + 1) π}{12}}) \Rightarrow R (x) = \frac{\prod_{k = 0}^{11} (x - e^{i \frac{(2 k + 1) π}{n}})}{x + 1}$
Commented by MJS last updated on 03/Nov/18

$\frac{x^{12} + 1}{x + 1} = x^{11} - x^{10} + x^{9} - x^{8} . . . + x - 1 + \frac{2}{x + 1}$
Commented by maxmathsup by imad last updated on 03/Nov/18

$t h a n k s s i r f o r t h i s r e m a k .$
Commented by rahul 19 last updated on 04/Nov/18
thanks sir!��
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Nov/18
	$S_1 =((1{(x^2 )^n_1 −1})/(x^2 −1)) now 1×(x^2 )^(n_1 −1) =x^(22) n_1 =12 S_2 =((1{x^n_2 −1})/(x−1)) x^(n_2 −1) =x^(11) n_2 =12 (S_1 /S_2 )=((((x^2 )^(12) −1)/(x^2 −1))/((x^(12) −1)/(x−1)))=(((x^(12) +1)(x^(12) −1)×x−1)/((x+1)(x−1)×(x^(12) −1))) =((x^(12) +1)/(x+1)) R=(−1)^(12) +1=2$
	$S_{1} = \frac{1 {{(x^{2})}^{n_{1}} - 1}}{x^{2} - 1}$ $n o w 1 \times {(x^{2})}^{n_{1} - 1} = x^{22} n_{1} = 12$ $S_{2} = \frac{1 {x^{n_{2}} - 1}}{x - 1}$ $x^{n_{2} - 1} = x^{11} n_{2} = 12$ $\frac{S_{1}}{S_{2}} = \frac{\frac{{(x^{2})}^{12} - 1}{x^{2} - 1}}{\frac{x^{12} - 1}{x - 1}} = \frac{(x^{12} + 1) (x^{12} - 1) \times x - 1}{(x + 1) (x - 1) \times (x^{12} - 1)}$ $= \frac{x^{12} + 1}{x + 1}$ $R = {(- 1)}^{12} + 1 = 2$
	Commented by rahul 19 last updated on 03/Nov/18

	$S i r, c o r r e c t a n s w e r i s$ $2 (1 + x^{2} + x^{4} + . . . . . + x^{10}) . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 03/Nov/18

	$i s i t p o s s i b l e . . . l e t o t h e r s o l v e t h e q u e s t i o n . . . i$ $d l n o t h i n k s o . . .$
	Commented by $@ty@m last updated on 03/Nov/18

	$x + 1 ∣ x^{12} + 1 ∣ x^{11} - x^{10} . . . . . - 1$ $x^{12} + x^{11}$ $- - - -$ $- x^{11} + 1$ $- x^{11} - x^{10}$ $- - - - - -$ $x^{10} + 1$ $. . . . . . . . .$ $. . . . . . . . .$ $- - - - - -$ $- x + 1$ $- x - 1$ $- - - - - -$ $2$ $- - - - - -$
	Commented by rahul 19 last updated on 04/Nov/18
	thank you sirs! ��
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum