<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] 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.
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