Find the remainder when x^(2006) −1 is divided by x^4 +x^3 +2x^2 +x+1.


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 113928 by ZiYangLee last updated on 16/Sep/20

$Find the remainder when x^{2006} - 1$ $is divided by x^{4} + x^{3} + 2 x^{2} + x + 1 .$
	Answered by MJS_new last updated on 16/Sep/20
	$x^4 +x^3 +2x^2 +x+1=(x^2 +1)(x^2 +x+1) the remaining fractions of ((x^(2n) −1)/((x^2 +1)(x^2 +x+1))) are 0 for n=6k ((x^2 −1)/((x^2 +1)(x^2 +x+1))) for n=6k+1 [2006=2(6×167+1)] −((x+2)/(x^2 +x+1)) for n=6k+2 ((2x)/(x^2 +1)) for n=6k+3 −((2x+1)/(x^2 +x+1)) for n=6k+4 ((x^3 +x−2)/((x^2 +1)(x^2 +x+1))) for n=6k+5$
	$x^{4} + x^{3} + 2 x^{2} + x + 1 = (x^{2} + 1) (x^{2} + x + 1)$ $the remaining fractions of$ $\frac{x^{2 n} - 1}{(x^{2} + 1) (x^{2} + x + 1)}$ $are$ $0 for n = 6 k$ $\frac{x^{2} - 1}{(x^{2} + 1) (x^{2} + x + 1)} for n = 6 k + 1 [2006 = 2 (6 \times 167 + 1)]$ $- \frac{x + 2}{x^{2} + x + 1} for n = 6 k + 2$ $\frac{2 x}{x^{2} + 1} for n = 6 k + 3$ $- \frac{2 x + 1}{x^{2} + x + 1} for n = 6 k + 4$ $\frac{x^{3} + x - 2}{(x^{2} + 1) (x^{2} + x + 1)} for n = 6 k + 5$
	Commented by 1549442205PVT last updated on 16/Sep/20

	$For x^{2006} - 1 \Rightarrow n = 1003 = 6 \times 167 + 1$
	Commented by MJS_new last updated on 16/Sep/20

	$you are right, thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum