<s>[INST] What is the remainder when $x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$ is divided by: $x^2-8x+12$. What is the remainder when: $$x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$$ is divided by: $$x^2-8x+12$$
$x^2-8x+12$ $\leftrightarrow (x-2)(x-6)$
This gives me the polynomial: $(x-2)(x-6)k(x) + r(x)$, where $r(x)$ is the remainder. 
The expression gives me the remainder when $x=2$ or $x = 6$
By replacing $x$ with $2$ into the original polynomial you get: $$x^{100} - 4x^{100}+3x^{100}-3x^{10}+12x^{10}-9x^{10}+3x^{2}-29x + 41 = 3x^2-29x + 41$$
By plugging in $2$ I get: $12-58 + 41 = -5$.
Hence $r(x)$ should be: $-5$.
This is apparently the wrong answer. What did I do wrong?
Thank you kindly for your help!
 [/INST] First note that $r(x)=ax+b$ for some $a,b\in\mathbb Z$
For $x=2$ you get $r(2)=-5$ (not $r(x)=-5$) $ \Rightarrow 2a+b=-5$. Do the same for $x=6$ to find r(6).Then find $a,b$.
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