What is $f(x)$ divided by $(x-a)$?

This is an exercise from Spivak's Calculus:

Prove that for any polynomial function $f$, and any number $a$, there is a polynomial function $g$, and a number $b$, such that $f(x)=(x-a)g(x)+b$ for all $x$. (The idea is simply to divide $(x-a)$ into $f(x)$ by long division, until a constant remainder is left. [. . .] A formal proof is possible by induction on the degree of $f$.

I did an adaptation of this question and decided to figure out precisely what $g(x)$ and $b$ are. Here is what I ended up with:

First, presume $f$ takes the form: $$f_n(x)=\sum_{u=0}^{n}c_ux^u$$

Then: $$f_n(x)=(x-a)\sum_{q=1}^{n}\sum_{z=q}^{n}a^{z-q}c_zx^{q-1}+\sum_{z=0}^{n}a^zc_z$$

I've had a really hard time figuring out a clean and nice way to show this is true. But, I'm pretty sure it's accurate. I can't quite show the extensive work in deriving this because it's a lot of synthetic division and long division, then noticing patterns among the remainders and the quotients. My manipulations of double sums are not that great, so I'd appreciate if anyone could either 1) show that I'm wrong (which is very possible) or 2) show how this is right, in a preferably elegant way.

One method is thus: Assuming that $b=\sum_{z=0}^{n}a^zc_z$, then $g(x)$ could be derived by 'basic' algebra: $$\sum_{u=0}^{n}c_ux^u=(x-a)g(x)+\sum_{z=0}^{n}a^zc_z \Rightarrow g(x)=\frac{\sum_{u=0}^{n}c_ux^u-a^{u}c_u}{(x-a)}$$ $$g(x)=\frac{\sum_{u=0}^{n}c_u(x^u-a^u)}{(x-a)}$$ $$\frac{x^u-a^u}{(x-a)}=\sum_{k=1}^{u}a^{u-k}x^{k-1}$$ $$\therefore g(x)=\sum_{u=0}^{n}\sum_{k=1}^{u}a^{u-k}c_ux^{k-1}$$

This seems to contradict what I just wrote, but it brings up another question: Is $$\sum_{u=0}^{n}\sum_{k=1}^{u}a^{u-k}c_ux^{k-1}=\sum_{q=1}^{n}\sum_{z=q}^{n}a^{z-q}c_zx^{q-1}\text{?}$$

I think it is, and I can show that thus:

First, change the right hand side: $\sum_{k=1}^{n}\sum_{u=k}^{n}a^{u-k}c_ux^{k-1}$. Then, make the substitution $e_{u,k}=a^{u-k}c_ux^{k-1}$

So, you have:

$$\sum_{u=0}^{n}\sum_{k=1}^{u}e_{u,k}=\sum_{k=1}^{0}e_{0,k}+\sum_{k=1}^{1}e_{1,k}+\sum_{k=1}^{2}e_{2,k}+\dots+\sum_{k=1}^{n-2}e_{(n-2),k}+\sum_{k=1}^{n-1}e_{(n-1),k}+\sum_{k=1}^{n}e_{n,k}$$ $$\sum_{u=0}^{n}\sum_{k=1}^{u}e_{u,k}=\sum_{k=1}^{n}e_{n,k}+\sum_{k=1}^{n-1}e_{(n-1),k}+\sum_{k=1}^{n-2}e_{(n-2),k}+\dots+\sum_{k=1}^{2}e_{2,k}+\sum_{k=1}^{1}e_{1,k}+\sum_{k=1}^{0}e_{0,k}$$

Each sum can be modified thusly:

$$\sum_{k=1}^{n-j}e_{(n-j),k}=\sum_{k=(1+j)}^{n}e_{(n-j),(k-j)}$$

So, by inspection, the two sums are equivalent.

I guess what my question now amounts to is thus: Are there any methods I'm simply not using that I could be using? And am I incorrect in any of my methods and statements above? Feel free to use a completely different approach to solve this, I'd like to see other approaches.

P.S. If you would like, feel free to provide a proof of the original problem by induction.

Here's a quick derivation along the lines suggested by @Pierre-YvesGaillard. It may be against the spirit of what Spivak intended since it avoids induction, but it is a different and simple way to go.

Suppose $f(x)$ is an $n$th degree polynomial. Then $$\begin{eqnarray*} f(x) &=& \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \\ &=& \sum_{k=1}^n \frac{f^{(k)}(a)}{k!} (x-a)^k + f(a) \\ &=& (x-a) \sum_{k=1}^n \frac{f^{(k)}(a)}{k!} (x-a)^{k-1} + f(a) \\ &=& (x-a) \underbrace{\sum_{k=0}^{n-1} \frac{f^{(k+1)}(a)}{(k+1)!} (x-a)^{k}}_{g(x)} + \underbrace{f(a)}_b. \end{eqnarray*}$$ As claimed, $g(x)$ is a polynomial (of degree $n-1$) and $b$ is a constant. We have used the fact that the Taylor expansion of a polynomial of degree $n$ terminates at the $n$th term, since $D^m x^n = 0$ for $m>n$. It is a good exercise to show that this $g$ and $b$ agree with your formula for $f$.

I recommend using the Iverson bracket when manipulating double sums (among other things). Here's a sample calculation: $$\begin{eqnarray*} \sum_{a=0}^n \sum_{b=0}^a &=& \sum_{a,b} [0\le a\le n] [0\le b \le a] \\ &=& \sum_{a,b} [0\le b\le a\le n] \\ &=& \sum_{a,b} [0 \le b \le n] [b \le a \le n] \\ &=& \sum_{b=0}^n \sum_{a=b}^n \end{eqnarray*}$$