1. Remainder Theorem

Find the remainder when $\displaystyle x^{100}+5$ is divided by $\displaystyle x^2-2x+1$

Anyone can help? Thanks

2. Originally Posted by acc100jt
Find the remainder when $\displaystyle x^{100}+5$ is divided by $\displaystyle x^2-2x+1$

Anyone can help? Thanks
$\displaystyle x^2-2x+1=(x-1)(x-1)$

The remainder theorem states that if a polynomial $\displaystyle f(x)$ is divided by $\displaystyle x-k$, the remainder is $\displaystyle r=f(k)$.

So, if I were to divide $\displaystyle x^{100}+5$ by $\displaystyle (x-1)$, this implies that the remainder is $\displaystyle (1)^{100}+5=6$. If I were to do it again...

I think. It's been a long time man.

3. i understand the first part. but what do u mean "do it again"?what is the dividend if i wan to divide again?

4. Originally Posted by acc100jt
i understand the first part. but what do u mean "do it again"?what is the dividend if i wan to divide again?
By the Division Algorithm

$\displaystyle f(x)=d(x)q(x)+r(x)$

The way to use this is

$\displaystyle \frac{f(x)-r(x)}{d(x)}=q(x)$

where $\displaystyle f(x)=x^{100}+5,r(x)=6,\text{ and },q(x)=\text{ the new function to be divided}$

5. this is what I've tried,

Let $\displaystyle f(x)=x^{100}+5$

$\displaystyle f(1)=6$

So
$\displaystyle f(x)=(x-1)Q(x)+6$....(1)
$\displaystyle f(x)-6=(x-1)Q(x)$
$\displaystyle x^{100}-1=(x-1)Q(x)$
$\displaystyle (x-1)(x^{99}+x^{98}+...+x+1)=(x-1)Q(x)$

Thus $\displaystyle Q(x)=x^{99}+x^{98}+...+x+1$

Since $\displaystyle Q(1)=100$
So $\displaystyle Q(x)=(x-1)P(x)+100$, substitute this into (1)

$\displaystyle f(x)=(x-1)[(x-1)P(x)+100]+6$
$\displaystyle f(x)=(x-1)^{2}P(x)+100x-94$

Hence the remainder is $\displaystyle 100x-94$

Am I correct?