The remainder obtained when the polynomial x+x^3 +x^9+x^27+x^81+x^2^243 is divided by x^2-1 is:
The answer is given as 6x but I dont know how to proceed.
Thanks
Shouldn't it be: $\displaystyle P(x)=x+x^3 +x^9+x^{27}+x^{81}+x^{243}$?
Anyway, when divided by $\displaystyle x^2-1=(x-1)(x+1)$:
$\displaystyle P(x)=[Q(x)](x-1)(x+1) + R(x)$, where Q(x) is the quotient, R(x) is the remainder. Its degree can be at most 1, so $\displaystyle R(x)=ax+b$ and
$\displaystyle P(x)=[Q(x)](x-1)(x+1) + ax+b$
Now
$\displaystyle P(1)=1+1^3+1^9+1^{27}+1^{81}+1^{243}=6$
$\displaystyle P(-1)=-1+(-1)^3+(-1)^9+(-1)^{27}+(-1)^{81}+(-1)^{243}=-6$
but also $\displaystyle P(1)=a+b$
$\displaystyle P(-1)=-a+b$
Now you've got 2 simultaneous equations, solving for a and b gives a=6, b=0 so R(x)=6x.