the remainder obtained when x^2006 is divided by x^2-1

Asked by amma95 | 14th Sep, 2009, 01:01: PM

Expert Answer:

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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