Long Division of Improper Rational Functions

# Long Division of Improper Rational Functions

We are about to look at a very important technique of integration known as Integration with Partial Fractions, however, we will first look at a not-so-calculus technique often necessary to apply the integration with partial fractions method. We begin with a few important definitions.

 Definition: If $f$ is a rational function, then the Degree of $f$ often denoted $\mathrm{deg} (f)$ is equal to the highest value exponent in $f$.

For example, if $f(x) = x^2 + x^3 - 2x$, then $\mathrm{deg} (f) = 3$ since $x^3$ is the term containing the highest exponent. In the function $g(x) = (x^2 + 1)^2 + x$, $\mathrm{deg} (g) = 4$ since when expanded, $g(x) = x^4 + 2x^2 + x + 1$.

 Definition: If $f(x) = \frac{P(x)}{Q(x)}$ is a rational function, and $\mathrm{deg} (P) ≥ \mathrm{deg} (Q)$, then we say that $f$ is an Improper Rational Function. Furthermore, if $\mathrm{deg} (P) < \mathrm{deg} (Q)$, then we say that $f$ is a Proper Rational Function.

The technique of long division will only apply to improper rational functions as we will see.

 Polynomial Long Division Algorithm: If $f(x) = \frac{P(x)}{Q(x)}$ where $\mathrm{deg}(P) ≥ \mathrm{deg}(Q)$, then $f(x) = S(x) + \frac{R(x)}{Q(x)}$ where $S(x)$ and $R(x)$ are polynomials and $\mathrm{deg} (R) < \mathrm{deg} (Q)$.

First, let's look at the example $f(x) = \frac{x^3 + 2x^2 + 1}{x - 1}$. We our now going to apply long division to this improper rational function. We note that $P(x) = x^3 + 2x^2 + 1$ and $Q(x) = x - 1$.

 Note: In our example, the degrees of each term in $P(x)$ and $Q(x)$ are descending. For this algorithm to work properly, it is best to reorder the terms of $P(x)$ and $Q(x)$ if they're not descending in degree.

Take the first term of $P$ and divide it by the first term of $Q(x)$. In our case, we get that $x^3$ divided by $x$ is $x^2$. Now multiply $x^2$ by the divisor $Q(x)$ and find the difference $P(x) - x^2Q(x)$ as follows:

(1)
\begin{align} \quad P(x) - x^2Q(x) = (x^3 + 2x^2 + 1) - x^2(x - 1) \\ \quad P(x) - x^2Q(x) = (x^3 + 2x^2 + 1) - (x^3 - x^2) \\ \quad P(x) - x^2Q(x) = 3x^2 + 1 \end{align}

Now take the first term of this new function, in our case, it is $3x^2$, and divide it by the first term in $Q(x)$, which of course is still $x$. We note that $3x^2$ divided by $x$ equals $3x$. Now take $[P(x) - x^2Q(x)]$ and subtract $3xQ(x)$ as follows:

(2)
\begin{align} \quad [P(x) - x^2Q(x)] - 3xQ(x) = (3x^2 + 1) - 3x(x - 1) \\ \quad [P(x) - x^2Q(x)] - 3xQ(x) = (3x^2 + 1) - (3x^2 - 3x) \\ \quad [P(x) - x^2Q(x)] - 3xQ(x) = 3x + 1 \end{align}

Now take the first term of this new function, in our case, it is $3x$, and divide it by the first term in $Q(x)$, which is $x$. We note that $3x$ divided by $x$ is $3$. Now take $[[P(x) - x^2Q(x)] - 3xQ(x)] - 3Q(x)$ as follows:

(3)
\begin{align} \quad [[P(x) - x^2Q(x)] - 3xQ(x)] - 3Q(x) = (3x + 1) - 3(x - 1) \\ \quad [[P(x) - x^2Q(x)] - 3xQ(x)] - 3Q(x) = (3x + 1) - (3x - 3) \\ \quad [[P(x) - x^2Q(x)] - 3xQ(x)] - 3Q(x) = 4 \end{align}

We now stop since the degree of this function is less than $\mathrm{deg}(Q) = 1$. We say that $R(x) = 4$, that is, our remainder is $4$, and we say that $S(x) = x^2 + 3x + 3$. Therefore:

(4)
\begin{align} \quad f(x) = \frac{x^3 + 2x^2 + 1}{x - 1} = x^2 + 3x + 3 + \frac{4}{x - 1} \end{align}