Problem source

Show that, when the polynomial ${\rm p} (x)$ is divided by $(x-a)$,  where $a$ is a real number, the remainder is ${\rm p}(a)$.
\begin{questionparts}
\item When the polynomial ${\rm p}(x)$ is divided by $x-1,\,x-2,\,x-3$ the remainders are 3,1,5 respectively. Given that $${\rm p}(x)=(x-1)(x-2)(x-3){\rm
q}(x)+{\rm r} (x),$$
 where ${\rm q}(x)$ and ${\rm r}(x)$ are polynomials with ${\rm r}(x)$ having degree less than three, find ${\rm r}(x)$.
\item Find a polynomial ${\rm P}(x)$ of degree $n+1$, where $n$ is a  given positive integer, such that for each integer $a$ satisfying $0\le a\le n$, the remainder when ${\rm P}_n(x)$ is divided by $x-a$ is $a$. 
\end{questionparts}

Solution source

Notice by polynomial division, we can write $p(x) = (x-a)q(x) + r(x)$ where degree $r(x) < 1$, ie $r(x)$ is a constant. Evaluating at $x = a$, we have $p(a) = (a-a)q(a) + r(a) = r(a)$. Therefore $r(a) = p(a)$ and since $r(x)$ is a constant, it is always $p(a)$.

\begin{questionparts}
\item $\,$
\begin{align*}
&& p(x) &= (x-1)(x-2)(x-3)q(x) + r(x) \\
&& p(1) &= r(1) = 3 \\
&& p(2) &= r(2) = 1 \\
&& p(3) &= r(3) = 5
\end{align*}

Therefore $r(x)$ is a polynomial of degree $2$ or less through $(1,3),(2,1), (3, 5)$ we can write this as

\begin{align*}
&& r(x) &= 5\frac{(x-1)(x-2)}{(3-1)(3-2)} + 1\frac{(x-1)(x-3)}{(2-1)(2-3)} + 3\frac{(x-2)(x-3)}{(1-2)(1-3)} \\
&&&= \frac52(x^2-3x+2)-(x^2-4x+3) + \frac32(x^2-5x+6) \\
&&&= 3x^2-11x+11
\end{align*}

\item Let $P_n(x) = x(x-1)\cdots(x-n) + x$, then $P_n(a) = a$

\end{questionparts}