IIT JEE , AIEEE Forum - Mathematics , Algebra

man111

(1) find the remainder when we divide x+x³+x⁹+x²⁷+x⁸¹+x²⁴³ is divided by x²-1

ans=(a)6x+1

(b)5x+1

(c)4x

(d)6x

allamraju

Let the given fun. be f(x).Then,we write,

f(x)=(x²-1)g(x)+R(x)=(x²-1)g(x)+(ax+b)

Now,put x=1 and -1 in above eqn.,we have,

a+b=6 and -a+b=-6 and so,a=6,b=0.

Hence the remainder is 6x.

hsbhatt

There is an interesting way to do this using the method of congruences:

$x^2 \equiv 1 \bmod{x^2-1} \Rightarrow x^{2n} \equiv 1 \bmod{(x^2-1)} \\ \\ \Rightarrow x^{2n+1} \equiv x \bmod{(x^2-1)} \\ \\ \text{Obviously} \ x \equiv x \bmod{(x^2-1)} \\ \\ \text{Hence, we get} \ x+x^3+x^9+x^{27}+x^{81}+x^{243} \equiv 6x \bmod {(x^2-1)} \\ \\ \text{In other words,} \\ \\ \ (x+x^3+x^9+x^{27}+x^{81}+x^{243}) \ \text{leaves a remainder 6x when divided by} \ (x^2-1) $

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Let the given fun. be f(x).Then,we write,

f(x)=(x2-1)g(x)+R(x)=(x2-1)g(x)+(ax+b)

Now,put x=1 and -1 in above eqn.,we have,

a+b=6 and -a+b=-6 and so,a=6,b=0.

Hence the remainder is 6x.

f(x)=(x²-1)g(x)+R(x)=(x²-1)g(x)+(ax+b)