What is the remainder when #x^1999# is divided by #(x^2-1)# ?

2 Answers
Apr 9, 2017

#x#

Explanation:

Note that:

#(x^2-1)(x^1997+x^1995+x^1993+...+x^3+x)#

#=x^2(x^1997+x^1995+x^1993+...+x^3+x)-(x^1997+x^1995+x^1993+...+x^3+x)#

#=(x^1999+color(red)(cancel(color(black)(x^1997)))+color(purple)(cancel(color(black)(x^1995)))+...+color(cyan)(cancel(color(black)(x^5)))+color(green)(cancel(color(black)(x^3))))-(color(red)(cancel(color(black)(x^1997)))+color(purple)(cancel(color(black)(x^1995)))+color(blue)(cancel(color(black)(x^1993)))+...+color(green)(cancel(color(black)(x^3)))+x)#

#=x^1999-x#

So:

#x^1999 = (x^2-1)(x^1997+x^1995+x^1993+...+x^3+x)+x#

That is:

#x^1999/(x^2-1) = x^1997+x^1995+x^1993+...+x^3+x#

with remainder #x#

The polynomial #x# is of degree less than the degree of #x^2-1#, so is the remainder we want.

Apr 9, 2017

#x#

Explanation:

Calling

#p_n(x)=x^n+c_(n-1)x^(n-1)+ cdots+c_0#

as a generic #n# degree polynomial we have.

#x^n = p_(n-2)(x^2-1) + ax+b# with #n ge 2#

so we have

#(-1)^n = a(-1)+b#

and

#1 = a+b#

solving we have

#a = (1-(-1)^n)/2# and #b=(1+(-1)^n)/2#

in the present case we have #n=1999# so

#a=1, b=0# and the remainder is #x#