Hello, can anyone help me with this question? Thanks in advance
Find the remainder when (x^100)+5 is divided by (x^2)-2x+1.
Hello,
The remainder has to be a polynomial of at most degree 1.
This means that it is in the form R(x)=ax+b, where a and b can be equal to 0.
Let's denote $\displaystyle P(x)=x^{100}+5$ and note that $\displaystyle x^2-2x+1=(x-1)^2$
What we'll do is studying the zeroes of the polynomials...
We have $\displaystyle P(x)=(x-1)^2Q(x)+R(x)$
where Q is the quotient.
If we let x=1, we have $\displaystyle P(1)=6=0*Q(1)+R(1)$
Thus $\displaystyle \boxed{R(1)=a+b=6}$
If we differentiate, we have $\displaystyle P'(x)={\color{red}2(x-1)Q(x)+(x-1)^2Q(x)}+R'(x)$
But $\displaystyle P'(x)=100x^{99}$ and $\displaystyle R'(x)=a$
Again, if we let x=1, the red part will be equal to 0.
Which gives $\displaystyle \boxed{1=a}$
Thus b=... and the remainder R is...