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?


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 $6$.
$$
(5x - 3)(x + 1) + 6 = 5x^2 + 5x - 3x - 3 + 6 = 5x^2 + 2x + 3 
$$
