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?


This is an easier way to do polynomial division, at least I think it is easier. 
Basically when we multiply polynomials (or any numbers really) we can sometimes do it in a table form, say if we wanted to multiply $x + 2$ and $x^2 + 3x + 4$ we might write a table like this: 
 $$
         \\ \begin{array} {c|c|c|c} 
         multiply&x^2 & 3x &4 \\ \hline
         x &x^3 &3x^2 &4x \\ \hline
         2 &2x^2 & 6x & 8
         \end{array}
         $$
And from this we get that the product of $x + 2$ and $x^2 + 3x + 4$ is $x^3  + 5x^2 + 10x + 8$. Now we want to do this in reverse. 
We know that the highest power term in the product ($2x^2$ in this case) must be the product of the highest powers in the factors. This might seem a bit strangely worded, but basically what I mean is that we know our quadratic (not trinomial) is the product of two linear (not binomial) factors, so our $2x^2$ term only comes from the product of the $x$ terms. So, we know when we do the table multiplication as above we have:
$$
         \\ \begin{array} {c|c|c} 
         divide& * &  \\ \hline
         x &\color{Red}{2x^2} &  \\ \hline
         1 &** &
         \end{array}
         $$
And now we know what the entry in $*$ must be: $2x$, and then we get $**$ also so we have
$$
         \\ \begin{array} {c|c|c} 
         divide& \color{Red}{2x} & ** \\ \hline
         x &{2x^2} & * \\ \hline
         1 & \color{Red}{2x}&
         \end{array}
         $$
And now, because in our product $2x^2 + 5x + 3$ we have $5x$, and the power of the term in $*$ is x, we must have that $2x + * = 5x$ so we get $ * = 3x$ so then $** = 3$ so our table is :
$$
         \\ \begin{array} {c|c|c} 
         divide& {2x} & \color{red}{3} \\ \hline
         x &{2x^2} & \color{red}{3x} \\ \hline
         1 & {2x}& *
         \end{array}
         $$
And lastle we see * = 3 which is what we want, so we get that $(x+1)(2x+3) = 2x^2 + 5x + 3$ so $\frac{2x^2 + 5x + 3} {x + 1} = 2x + 3$.
This works in general for any order polynomial, and if one polynomial doesn't divide the other, you can easily get the remainder from this algorithm too.
