the remainder obtained when x^2006 is divided by x^2-1
Asked by amma95 | 14th Sep, 2009, 01:01: PM
Applying division algorithm when x^2006 is divided by x^2-1
x^2006 = (x^2-1)p(x)+r(x) ,where r(x) is a polynomial of degree atmost 1 ......(i)
let r(x)= Ax+B
(i) becomes
x^2006 = (x^2-1)p(x)+ Ax+B
substituting x=1 and x=-1 we get
A+B=1; A-B = 1 Solving we have A=0 B=0 so r(x) =1 option D isthe answer
Answered by | 21st Oct, 2009, 11:02: AM
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