<s>[INST] How to divide polynomials? The title says it all : How to divide polynomials?
I don't understand the method below taught in my school. 
Would any of you mind explaining it or even better, suggest an alternative way to solve this?

 [/INST] Here is another method that you asked for.  Its main advantage is that you can easily do this in a text editor, which helps immensely in accurately tracking your work.  It is completely done with algebraic manipulation.
So, with $A$ as the highest term in numerator and $B$ as the sum of all remaining terms.   And $C$ as the highest term in denominator, and $D$ is the remaining terms.  The goal is to eliminate $A$ and get an answer in terms of $m + \frac{r}{C + D}$.  In other words, $m$ is a multiple of the denominator, and $r$ is the remainder.  So you start with two things being divided.
Here we justify an algebraic procedure.  Imagine some example:
$$
\color{red}{A = 4x^2}
$$
$$
\color{green}{B = 3x + 5}
$$
$$
\color{blue}{C = 2x^2}
$$
$$
\color{purple}{D = 9x - 7}
$$
So the polynomial being divided is like this, at a high level.
So ignore their values and see what happens when we perform one step:
$$
\frac{A + B}{C + D} =
$$
And note that you can add and subtract a ratio of the highest terms $\frac{A}{C}$ without changing the answer.
$$
\frac{A}{C} - \frac{A}{C} + \frac{A + B}{C + D} =
$$
And you can multiply and divide by the denominator $C + D$ without changing the answer as well.  Specifically, you can do this to the subtracted term.  It just happens that $A$, which represents the highest term will always cancel out when you simplify this.  This is actually the only thing you need to remember.  You can derive everything else with this.
$$
\frac{A}{C} - \frac{A}{C}*\frac{C+D}{C+D} + \frac{A + B}{C + D} =
$$
If you start there, you can usually jump right to that step in your head.
So now, we can figure out what it simplifies to, just to see what happens with $A$:
$$
\frac{A}{C} + \frac{-AC -AD}{CC + CD} + \frac{AC + BC}{CC + DC} =
$$
$$
\frac{A}{C} + \frac{BC - AD}{C(C + D)}
$$
So, let's apply it to the example values:
$$
\frac{\color{red}{4x^2} + \color{green}{(3x + 5)}}{\color{blue}{2x^2}+ \color{purple}{(9x - 7)}} =
$$
$$
\frac{\color{red}{4x^2}}{\color{blue}{2x^2}} + \frac{(\color{green}{3x + 5})(\color{blue}{2x^2}) - (\color{red}{4x^2})(\color{purple}{9x-7})}{(\color{blue}{2x^2})(\color{blue}{2x^2}+ \color{purple}{(9x - 7)})} =
$$
$$
\frac{4x^2}{2x^2} + \frac{6x^3 + 10x^2 - 36x^3 + 28x^2}{(2x^2)(2x^2+ (9x - 7))} =
$$
$$
\frac{4x^2}{2x^2} + \frac{3x + 5 - 18x + 14}{2x^2+ (9x - 7)} =
$$
$$
2 + \frac{-15x + 19}{2x^2+ (9x - 7)}
$$
So, the current answer is $2$ plus remainder $(-15x+19)$.  If $\frac{-15x}{2x^2} $ has a positive $x$ exponent, then we apply this method to the remainder to get more terms.  Let's check the answer:
$$
2(2x^2+ (9x - 7)) + (-15x + 19) =
$$
$$
(4x^2+ 18x - 14) + (-15x + 19) =
$$
$$
\color{red}{4x^2} + \color{green}{3x +5}
$$
Which is our numerator $\color{red}{A} + \color{green}{B}$.
Let's apply this to a different polynomial that requires more than one reduction:
$$
\frac{\color{red}{5x^2} + \color{green}{2x + 3}}{\color{blue}{x} + \color{purple}{1}} =
$$
$$
\frac{5x^2}{x} + \frac{(2x + 3)x - 5x^2}{x(x + 1)} =
$$
$$
5x + \frac{2x + 3 - 5x}{x + 1} =
$$
$$
5x + \frac{\color{red}{-3x} + \color{green}{3}}{\color{blue}{x} + \color{purple}{1}} =
$$
$$
5x  - 3 + \frac{3x + 3x}{x(x + 1)} =
$$
$$
5x  - 3 + \frac{6}{x + 1} =
$$
So, let's check that the answer is $5x - 3$ with remainder $6$.
$$
(5x - 3)(x + 1) + 6 = 5x^2 + 5x - 3x - 3 + 6 = 5x^2 + 2x + 3 
$$
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