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: