Polynomial question 2

**champrock** · April 12th 2009, 03:20 AM

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

**pinkparrot** · April 12th 2009, 04:37 AM

Shouldn't it be: $P(x)=x+x^3 +x^9+x^{27}+x^{81}+x^{243}$ ?

Anyway, when divided by $x^2-1=(x-1)(x+1)$ :

$P(x)=[Q(x)](x-1)(x+1) + R(x)$ , where Q(x) is the quotient, R(x) is the remainder. Its degree can be at most 1, so $R(x)=ax+b$ and

$P(x)=[Q(x)](x-1)(x+1) + ax+b$
Now
$P(1)=1+1^3+1^9+1^{27}+1^{81}+1^{243}=6$
$P(-1)=-1+(-1)^3+(-1)^9+(-1)^{27}+(-1)^{81}+(-1)^{243}=-6$

but also $P(1)=a+b$
$P(-1)=-a+b$

Now you've got 2 simultaneous equations, solving for a and b gives a=6, b=0 so R(x)=6x.

**champrock** · April 12th 2009, 08:56 AM

nopes . x^2^243 only

anyways, will try with your method and c if i can get an answer.

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