A polynomial division is a method to rewrite a fraction $\frac{p(x)}{q(x)}$ of polynomials as

$$\frac{p(x)}{q(x)} = h(x) + r(x)\text,$$ where $r$ is a polynomial of degree less than $q$. This $r$ is the remainder. We will first discuss polynomial divisions without remainder, that is, $r = 0$.

Suppose we wish to simplify the fraction $\frac{x^3-1}{x-1}$. We prepare our polynomial division by writing this as follows:

$$\begin{array}{lclcr}x-1 &/ &\begin{array}[t]{lr}x^3-1\end{array} &\backslash & \end{array}$$ The final answer will appear on the right-hand side of the second slash and we will work out the details of the polynomial division below the $x^3-1$. In every step of the so-called long division we will subtract multiples of $x-1$ from $x^3-1$ until we are left with a small remainder $r$. A small remainder here means that the highest power of $x$ should be as small as possible.

In the first step of the long division we want to subtract a multiple of $x-1$ from $x^3-1$ which makes the term with the highest power, $x^3$, disappear. This means we have to subtract $x^2 \left( x-1 \right)$ from $x^3-1$. If you would subtract $-x^3 \left(x - 1 \right)$ from $x^3-1$ the $x^3$ would disappear as well, but this would introduce a new term with $x^4$. This is undesirable since we want to reduce the degree of the remainder and not increase it. We write down $x^2$ on the right side and $x^2(x-1)=x^3-x^2$ below $x^3-1$. The first step hence looks as follows:

$$\begin{array}{lclcr}x-1 &/ &\begin{array}[t]{lr}x^3-1\\{\blue{x^3-x^2}}&-\\ \hline \end{array} &\backslash & {\blue{x^2}}\end{array}$$ We now subtract $x^3-x^2$ from $x^3-1$ and the highest power $x^3$ indeed cancels: $$\begin{array}{lclcr}x-1 &/ &\begin{array}[t]{lr}x^3-1\\{\blue{x^3-x^2}}&-\\ \hline x^2-1\end{array} &\backslash & {\blue{x^2}}\end{array}$$ De remainder now is $x^2+x+1$. To cancel out the $x^2$ we subtract $x\left(x-1\right)$ in the second step: $$\begin{array}{lclcr}x-1 &/ &\begin{array}[t]{lr}x^3-1\\{\blue{x^3-x^2}}&-\\ \hline x^2-1\\{\blue{x^2-x}}&-\\ \hline x-1\end{array} &\backslash & {\blue{x^2+x}}\end{array}$$ The remainder now precisely is $x-1$. The completed long division therefore looks like this: $$\begin{array}{lclcr}x-1 &/ &\begin{array}[t]{lr}x^3-1\\{\blue{x^3-x^2}}&-\\ \hline x^2-1\\{\blue{x^2-x}}&-\\ \hline x-1\\{\blue{x-1}}&-\\ \hline 0\end{array} &\backslash & {\blue{x^2+x+1}}\end{array}$$
This means that $\frac{x^3-1}{x-1}=x^2+x+1$ and if you calculate $(x^2+x+1)(x-1)$ this indeed equals $x^3-1$.

In the previous example we ended up without a remainder, but unfortunately this is not always the case. If we for example do a long division of $\frac{x^3+2}{x+1}$ the remainder is $1$:

$$\begin{array}{lclcr}x+1 &/ &\begin{array}[t]{lr}x^3+2\\{\blue{x^3+x^2}}&-\\ \hline 2-x^2\\{\blue{-x^2-x}}&-\\ \hline x+2\\{\blue{x+1}}&-\\ \hline 1\end{array} &\backslash & {\blue{x^2-x+1}}\end{array}$$ This means that $$x^3 + 2 = (x^2-x+1)(x+1) + 1\text,$$ so $$\frac{x^3+2}{x+1} = x+1 + \frac{1}{x+1} \text.$$

In the next examples more long divisions are shown: