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.
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?