Let $f = 2018X^{2019} - 2019X^{2018} + 1, f \in \mathbb{R}[X]$. Find the quotient and the remainder when dividing $f$ to $(X-1)^2$.
The answer to this problem is that $f = (X-1)^2(2018X^{2017} + 2017X^{2016} + \cdots + 2X + 1)$. My question is: is there any way of obtaining this result in an elegant way, rather than by simply brute-forcing the long division and spotting the solution?
Not exactly brute force, but the neat answer allows us to use the synthetic division to find the answer.
Sorry for the picture, I'll try to format them with MathJax, but it might take a while.
First, by evaluating $f(1)$, we can find the remainder of the division by $(X-1)$
$$f(1) = 2018\cdot 1^{2019} - 2019 \cdot 1^{2018} + 1 = 0$$
Now dividing $f(X)$ by $(X-1)$
$$
\begin{array}{r|rrrrrrrr}
1&2018&-2019&0&0&0&\dots&0&1\\
&&2018&-1&-1&-1&\dots&-1&-1\\\hline
&2018&-1&-1&-1&-1&\dots&-1&0
\end{array}$$
$$f(X) = (X-1) (2018X^{2018} - X^{2017} - X^{2016} - X^{2015} - \dots - X^1 - 1)$$
This new polynomial is also divisible by $(X-1)$. Let's do the synthetic division again.
$$
\begin{array}{r|rrrrrrrr}
1&2018&-1&-1&-1&-1&\dots&-1&-1\\
&&2018&2017&2016&2015&\dots&2&1\\\hline
&2018&2017&2016&2015&2014&\dots&1&0
\end{array}$$
Giving us the desired answer
$$f(X) = (X-1)^2(2018X^{2017} + 2017X^{2016} + 2016X^{2015} + \dots + 2X^1 + 1)$$
$$ \begin{align} f(X) &= 2018X^{2019} - 2019X^{2018} + 1 \\ &= 2018 X^{2018}\left(X - 1\right) - \left(X^{2018} - 1\right) \\ &= \left(X - 1\right)2018 X^{2018} - \left(X - 1\right)\left(X^{2017}+ X^{2016} + \dots + 1\right) \\ &= (X-1)\left(\left(X^{2018} - X^{2017}\right) + \left(X^{2018} - X^{2016}\right)+\dots+\left(X^{2018} - 1\right)\right) \\ &= (X-1)^2\left(X^{2017} + X^{2016}(X+1) + X^{2015}\left(X^2+X+1\right)+\dots \\ +X^0\cdot\left(X^{2017}+X^{2016}+\dots+1\right)\right) \\ &= \dots \end{align} $$
