Why does remainder theorem works? If we divide the polynomial $f(x)$ by $(x - a)$, we get a remainder $r$ and a quotient $g(x)$. So, $f(x)$ can be written as,
$$f(x) = (x - a) g(x) + r$$
If $x=a$ then,
$f(a) = (a - a) g(x) + r$
or, $f(a) = r$.
It is a clear proof of the remainder theorem. But unfortunately, I don't find it intuitive.
Can you show me any "intuitive" way to prove the remainder theorem?
 A: If it helps, a nice way to think about it is that $(x-a)$ is a factor of $f(x)-f(a)$. This is often called the 'factor theorem'.
If this is not unconvincing, note that plugging in $x=a$ gives $f(x)-f(a) = 0$ for obvious reasons, so $x=a$ is a root of the polynomial $g(x) = f(x)-f(a)$, which we know means that $(x-a)$ is a factor of $f(x)-f(a)$. Or else, you can see more directly by observing that $x^k - a^k = (x-a)(x^{k-1} + x^{k-2}a + \dots + a^{k-1})$, so the difference of each $k$th power term in $f(x)$ and $f(a)$ will be divisible by $(x-a)$.
In any case, if $x-a$ divides $f(x)-f(a)$, then $f(x)-f(a) = (x-a)h(x)$, where $h(x)$ is the other factors. Therefore, $f(x) = (x-a)h(x) + f(a)$, so the remainder is indeed $f(a)$.
If you want more detail you will need to ask for it.
A: Write
$$f(x) = c_nx^n + c_{n - 1}x^{n - 1} + \dots + c_1x +c_0$$
You want a polynomial $g(x) = d_{n - 1}x^{n - 1} + \dots + d_1x + d_0$ and a number $r \in \mathbb{R}$ such that $f(x) = (x - a)g(x) + r$, that is,
\begin{equation}\label{equation}
c_nx^n + c_{n - 1}x^{n - 1} + \dots + c_1x +c_0 = (x - a)(d_{n - 1}x^{n - 1} + \dots + d_1x + d_0) + r
\end{equation}
The polynomials in the left and right hand side are equal if, and only if they have the same coefficients.
The coefficient of $x^n$ in the right-hand side is $d_{n - 1}$, so we must have $d_{n - 1} = c_n$.
If $1 \leq k \leq n - 1$, the coefficient of $x^k$ in the right hand side is $d_{k - 1} - ad_{k}$, so we have $d_{k - 1} - ad_{k} = c_k$.
Finally, the coefficient of $x^0$ in the right hand side is $r - ad_0$, so that we have $r - ad_0 = c_0$.
Using the equations
$$d_{n - 1} = c_n$$
$$d_{k - 1} - ad_{k} = c_k \text{ for }1 \leq k \leq n - 1$$
$$r - ad_0 = c_0$$
We can find the unique polynomial $g(x)$ and number $r \in \mathbb{R}$ with
$$f(x) = (x - a)g(x) + r$$
