\(p(x)=(x-1)(x-2)(x-3)\\ r(x) = f(x) - p(x)q(x),~\text{ where }q(x) \text{ is the quotient of }f(x) \text{ and }p(x)\\ \text{note that }p(1)=p(2)=p(3) = 0\\ r(1) = f(1) - p(1)q(1) = f(1) = 2\\ r(2) = f(2) = 3\\ r(3) = f(3) = 5\)
\(\text{we have 3 equations so we are looking for a polynomial with 3 degrees of freedom}\\ \text{i.e. a 2nd degree polynomial }a x^2 + b x+ c\\ \text{plugging things in we get 3 equations}\\ a+b+c = 2\\ 4a+2b+c=3\\ 9a+3b+c=5\\ \text{This can be solved to obtain }\\ a=\frac{1}{2},b= -\frac{1}{2},c=2, \text{ and thus the remainder polynomial is}\\ r(x) = \dfrac 1 2(x^2 - x + 4)\)
.
) This is for any number, not just 10, of course.
