3. Rational fractions

A rational fraction is fraction where both the numerator and denominator are polynomials. Expressions of this type appear very frequently in science and mathematics, since polynomials are often used to approximate the behaviour of other functions over a range of values.

After completing this chapter you should be able to:

• distinguish between proper and improper rational fractions

• fully decompose rational fractions into simpler “partial fraction” form

3.1. Partial fractions

Composing fractions means putting everything together over a common denominator. For example:

(3.1)\[\frac{2}{x+1}+\frac{3}{x+2} = \frac{2(x+2)+3(x+1)}{(x+1)(x+2)}=\frac{5x+7}{x^2+3x+2}\]

Decomposition is the reverse process, i.e taking the fraction apart. The simpler fractions that arise from this process are called partial fractions. Although modern computer software or online tools such as Wolfram Alpha can do this for us, it may be helpful to understand a little about how it works.

Our main decomposition technique involves “guessing” the form by spotting common patterns. For example, we may observe that a product of factors in the denominator is obtained by adding fractions involving those factors, as in example (3.1). This may lead us to anticipate a result of the following form, where constants \(A\) and \(B\) are unknown:

(3.2)\[\frac{5x+7}{(x+1)(x+2)} = \frac{A}{x+1}+\frac{B}{x+2}\]

We can find the unknown values by comparing the target fraction to our guess. We do this by composing our guess and matching the numerators. In our example case we obtain the following by composing the right hand side:

(3.3)\[\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}=\frac{(A+B)x+(2A+B)}{(x+1)(x+2)}\]

There are two techniques that we may then use for matching the numerators:

Equating coefficients
Comparing coefficients of each power of \(x\) to our target fraction gives conditions for the unknown constants. In this case,

\[\begin{equation*} (A+B)x+(2A+B) = 5x+7 \quad \implies \quad \begin{array}{ll}A+B&=5\\ 2A+B&=7\end{array} \end{equation*}\]

The unique values of \(A\) and \(B\) satisfying both relationships are \(A=2,\) \(B=3\).

Exercise 3.1

Find the values of \(A,B,C,D\) that make the following relationship true:

\[\begin{equation*} \frac{x^3}{(x-1)^2}= Ax+B + \frac{C}{(x-1)}+\frac{D}{(x-1)^2} \end{equation*}\]

Solution

Composing the RHS gives

\[\begin{equation*} Ax+B + \frac{C}{(x-1)}+\frac{D}{(x-1)^2}=\frac{(Ax+B)(x-1)^2+C(x-1)+D}{(x-1)^2} \end{equation*}\]

Expanding the numerator gives

\[\begin{equation*} Ax^3+(B-2A)x^2+(A-2B+C)x+(B-C+D) \end{equation*}\]

For terms to match with \(x^3\) we require \(A=1\), \(B=2\), \(C=3\), \(D=1\).

Substituting values
Since the two polynomials should be equal for all values of \(x\) we can choose values that give the coefficients directly. In this case,

\[\begin{equation*} A(x+2)+B(x+1)=5x+7 \end{equation*}\]

• choosing \(x=-2\) eliminates the first factor and gives \(B=3\)

• choosing \(x=-1\) eliminates the second factor and gives \(A=2\)

Exercise 3.2

Find the values of \(A,B,C\) that make the following relationship true:

\[\begin{equation*} \frac{2}{(x^2-1)(x+2)} =\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x+2} \end{equation*}\]

Solution

Composing the RHS gives

\[\begin{equation*} \frac{A(x-1)(x+2)+B(x+1)(x+2)+C(x+1)(x-1)}{(x^2-1)(x+2)} \end{equation*}\]

• putting \(x=-1\) gives \(A=-1\),

• putting \(x=1\) gives \(B=1/3\),

• putting \(x=-2\) gives \(C=2/3\).

3.2. Proper fractions

A proper rational fraction is one where the degree of the numerator is less than the degree of the denominator.

Exercise 3.3

Which of the following are proper fractions?
(a) \(\displaystyle \frac{x}{x^2+1}, \qquad\) (b) \(\displaystyle \frac{x^2}{x^2+1}\)

Solution

(a) is a proper fraction since the numerator is degree 1 and the denominator is degree 2

(b) is not a proper fraction since the numerator and denominator are both degree 2

Any distinct factors appearing in the denominator of a proper fraction can be separated. For each fraction in the decomposition the degree of the numerator should be less than the degree of the denominator.

Examples

\[\begin{align*} \frac{3}{(x+2)(x+3)}&=\frac{A}{x+2}-\frac{B}{x+3}\\ \frac{x^2}{(x-1)(x^2+1)}&=\frac{A}{x-1}+\frac{Bx+C}{x^2+1} \end{align*}\]

Exercise 3.4

Determine which of the following fractions can be further decomposed, and find the decomposition for those that can:
a. \(\displaystyle\frac{x}{x^2-1}\quad\) b. \(\displaystyle\frac{1}{x^2-1}\quad\) c. \(\displaystyle \frac{x}{x^2+1}\quad\)

Solution

(a) \(\displaystyle\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\)
(b) \(\displaystyle\frac{1}{2(x-1)}-\frac{1}{2(x+1)}\)
(c) Cannot be further decomposed because the denominator cannot be factorised.

Repeated factors
Sometimes you may encounter a fraction featuring a repeated factor on the denominator. If we natively attempt to treat these the same way that we handled the case of distinct factors we will run into difficulties. For example, the following result is nonsense and cannot be satisfied for any values of the constants :

(3.4)\[\begin{equation} \frac{1}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{x+1} \end{equation}\]

Try it!

The combined fraction on the right is equal to

\[\begin{equation*} \frac{A+B}{x+1} \equiv \frac{(A+B)(x+1)}{(x+1)^2} \end{equation*}\]

There is no combination of \(A,B\) values that gives a non-zero constant on the numerator.

In fact, this fraction cannot be further decomposed. On the other hand, the following proper fraction which also features a repeated factor on the denominator can be decomposed as shown:

\[\begin{equation*} \frac{3x-2}{(x-1)^2} = \frac{3(x-1)+1}{(x-1)^2}=\frac{3}{x-1}+\frac{1}{(x-1)^2}. \end{equation*}\]

In general, for each repeated factor in the denominator with multiplicity \(k\) we can decompose it as a sum of \(k\) rational fractions, each containing a different power of the factor.

For example,

(3.5)\[\begin{align} \frac{x}{(x+1)^3}&=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}\\ \frac{x^3}{(x^2+1)^2}& = \frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2} \end{align}\]

Exercise 3.5

Determine which of the following fractions can be further decomposed, and find the decomposition for those that can:
a. \(\displaystyle\frac{x}{(x-1)^2}\quad\) b. \(\displaystyle\frac{x}{(x^2+1)^2}\quad\) c. \(\displaystyle\frac{x^2}{(x^2+1)^2}\quad\)

Solution

(a) \(\displaystyle\frac{1}{x-1}+\frac{1}{(x-1)^2}\)
(b) Cannot be further decomposed
(c) \(\displaystyle\frac{1}{x^2+1}-\frac{1}{(x^2+1)^2}\)

It is essential that you make sure the denominator is factorised as far as possible before doing the decomposition. For example

(3.6)\[\begin{equation} \frac{x^2}{(x^2-1)^2}=\frac{x^2}{(x-1)^2(x+1)^2} \end{equation}\]

Therefore the decomposition of this fraction has the form

(3.7)\[\begin{equation} \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2} \end{equation}\]

3.3. Improper fractions

An improper rational fraction is one where the degree of the numerator is greater than or equal to the degree of the denominator.

In that case we can split off a polynomial of degree \((m-n)\), where \(m,n\) are the degrees of the numerator and denominator, respectively.

The remaining proper fraction can then be further decomposed.

Examples

(3.8)\[\begin{equation} \frac{2x+3}{5x+2} =A+\frac{B}{5x+2} \end{equation}\]

explanation

Here we have a ratio of two polynomials of the same order. The decomposition will feature a degree 0 polynomial, i.e. a constant.

\[\begin{equation*} \frac{2x+3}{5x+2} \quad \text{is} \quad \frac{\text{degree 1}}{\text{degree 1}} \quad \implies \quad (m-n)=0 \end{equation*}\]

(3.9)\[\begin{equation} \frac{x^2}{5x+2} =Ax+B+\frac{C}{5x+2} \end{equation}\]

explanation

Here we have a ratio of two polynomials which differ by a single order. The decomposition will feature a degree 1 polynomial, i.e of the form \(Ax+B\).

\[\begin{equation*} \frac{x^2}{5x+2} \quad \text{is} \quad \frac{\text{degree 2}}{\text{degree 1}} \quad \implies \quad (m-n)=1 \end{equation*}\]

The constants in each case can be found by the method of equating coefficients that we introduced in this chapter. You should find that:

(3.10)\[\begin{align} \frac{2x+3}{5x+2} &=\frac{2}{5}+\frac{11/5}{5x+2}\\ \frac{x^2}{5x+2} &=\frac{1}{5}x-\frac{2}{25}+\frac{4/25}{5x+2} \end{align}\]

Note: we can also find these results by polynomial long division:

\( \require{enclose} \begin{array}{r r} &\frac{2}{5} \\ 5x+2 & \enclose{longdiv}{2x+3} \\ & 2x+\frac{4}{5}\\ \hline & \frac{11}{5} \end{array} \)

\( \require{enclose} \begin{array}{r r} &\frac{1}{5}x-\frac{2}{25} \\ 5x+2 & \enclose{longdiv}{x^2\phantom{+\frac{1}{5}x-\frac{2}{5}}} \\ & x^2+\frac{2}{5}x \phantom{+\frac{4}{25}}\\ \hline & \phantom{x^2}-\frac{2}{5}x \phantom{+\frac{4}{25}}\\ & \phantom{x^2}-\frac{2}{5}x-\frac{4}{5}\\ \hline & \frac{4}{25} \end{array} \)

Exercise 3.6

Decompose the following as far as possible: \(\displaystyle\frac{x^3}{x^2-1}\)

Solution

\(\displaystyle x+\frac{1}{2(x-1)}+\frac{1}{2(1+x)}\)