Polynomial, Rational and Irrational Functions

Lessons

Senior 4 · Mathematics

Polynomial, Rational and Irrational Functions

5 lessons

Lesson 5 of 5

Partial fractions

25 min

Running addition backwards

Adding two fractions is routine:

\frac{2}{x - 1} + \frac{3}{x + 2} = \frac{2(x + 2) + 3(x - 1)}{(x - 1)(x + 2)} = \frac{5x + 1}{(x - 1)(x + 2)} .

Going the other way, from $\frac{5x + 1}{(x - 1)(x + 2)}$ back to the simple pieces, is called splitting into partial fractions. The simple pieces are much easier to work with: each one has its own asymptote and its own behaviour, a long sum such as $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots$ collapses to a few terms, and in S5 and S6 integration of rational functions depends on it.

The solid curve $\frac{5x + 1}{(x - 1)(x + 2)}$ is the sum of the dashed curves $\frac{2}{x - 1}$ (orange) and $\frac{3}{x + 2}$ (green). Near $x = 1$ it behaves like $\frac{2}{x - 1}$ ; near $x = - 2$ like $\frac{3}{x + 2}$ .

Definition

Proper rational function: $\frac{P(x)}{Q(x)}$ with $\deg P \lt \deg Q$ . If $\deg P \ge \deg Q$ it is improper, and polynomial division writes it as a polynomial plus a proper fraction.

Irreducible quadratic: a quadratic $a x^{2} + b x + c$ with $\Delta = b^2 - 4ac \lt 0$ ; it has no real factors (for example $x^{2} + 1$ , $x^{2} + x + 1$ ).

Partial fractions of a proper rational function: its expression as a sum of fractions whose denominators are the factors of $Q (x)$ .

Important

The form of the partial fractions (for a proper fraction, after factorising the denominator completely)

Factor in the denominator	Partial fractions it gives
distinct linear factor $a x + b$	$\frac{A}{ax + b}$
repeated linear factor $(a x + b)^{2}$	$\frac{A}{ax + b} + \frac{B}{(ax + b)^2}$
irreducible quadratic $a x^{2} + b x + c$	$\frac{Ax + B}{ax^2 + bx + c}$

To find the constants, multiply both sides by the denominator to get an identity, then (1) put in the roots of the factors, and (2) compare coefficients for whatever is left.

Worked example 1: distinct linear factors

Express in partial fractions (a) $\frac{5x + 1}{(x - 1)(x + 2)}$ (b) $\frac{x^2 + 2x - 1}{x(x - 1)(x + 1)}$ .

Solution

$\frac{5x + 1}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$ , so $5 x + 1 = A (x + 2) + B (x - 1)$ . Put $x = 1$ : $6 = 3 A$ , $A = 2$ . Put $x = - 2$ : $- 9 = - 3 B$ , $B = 3$ . Hence $\frac{2}{x - 1} + \frac{3}{x + 2}$ .
$x^{2} + 2 x - 1 = A (x - 1) (x + 1) + B x (x + 1) + C x (x - 1)$ . $x = 0$ : $- 1 = - A$ , $A = 1$ . $x = 1$ : $2 = 2 B$ , $B = 1$ . $x = - 1$ : $- 2 = 2 C$ , $C = - 1$ . Hence $\frac{1}{x} + \frac{1}{x - 1} - \frac{1}{x + 1}$ .
Check at $x = 2$ : left side $\frac{7}{6}$ ; right side $\frac{1}{2} + 1 - \frac{1}{3} = \frac{7}{6}$ ✓.

Tip

Cover-up rule for a distinct linear factor $x - a$ : cover that factor in the denominator and put $x = a$ in what is left. For $\frac{5x + 1}{(x - 1)(x + 2)}$ , covering $x - 1$ and putting $x = 1$ gives $A = \frac{6}{3} = 2$ at a glance. It works only for distinct linear factors.

Worked example 2: a repeated factor

Express $\frac{3x^2 + x}{(x - 1)(x + 1)^2}$ in partial fractions.

Solution Write $\frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}$ . Then

3 x^{2} + x = A (x + 1)^{2} + B (x - 1) (x + 1) + C (x - 1) .

$x = 1$ : $4 = 4 A$ , so $A = 1$ . $x = - 1$ : $2 = - 2 C$ , so $C = - 1$ . Coefficients of $x^{2}$ : $3 = A + B$ , so $B = 2$ . Hence

\frac{3x^2 + x}{(x - 1)(x + 1)^2} = \frac{1}{x - 1} + \frac{2}{x + 1} - \frac{1}{(x + 1)^2} .

Check at $x = 0$ : left side $0$ ; right side $- 1 + 2 - 1 = 0$ ✓.

Warning

A repeated factor $(x + 1)^{2}$ needs both $\frac{B}{x + 1}$ and $\frac{C}{(x + 1)^2}$ . Writing only $\frac{C}{(x + 1)^2}$ leaves too few constants, and the identity then has no solution.

Worked example 3: an irreducible quadratic factor

Express $\frac{x^2 + 4x + 1}{(x - 1)(x^2 + 1)}$ in partial fractions.

Solution $x^{2} + 1$ has $\Delta = -4 \lt 0$ , so write $\frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$ :

x^{2} + 4 x + 1 = A (x^{2} + 1) + (B x + C) (x - 1) .

$x = 1$ : $6 = 2 A$ , $A = 3$ . $x^{2}$ : $1 = A + B$ , $B = - 2$ . Constant: $1 = A - C$ , $C = 2$ . Hence

\frac{x^2 + 4x + 1}{(x - 1)(x^2 + 1)} = \frac{3}{x - 1} + \frac{-2x + 2}{x^2 + 1} = \frac{3}{x - 1} - \frac{2x - 2}{x^2 + 1} .

Check at $x = 0$ : $\frac{1}{-1} = -1$ and $- 3 + 2 = - 1$ ✓.

Worked example 4: an improper fraction

Express $\frac{x^3 + 2x^2 - x + 1}{x^2 + x - 2}$ as a polynomial plus partial fractions.

Solution The numerator has the higher degree, so divide first: $x^{3} + 2 x^{2} - x + 1 = (x^{2} + x - 2) (x + 1) + 3$ . Then, with $x^{2} + x - 2 = (x - 1) (x + 2)$ and the cover-up rule,

\frac{x^3 + 2x^2 - x + 1}{x^2 + x - 2} = x + 1 + \frac{3}{(x - 1)(x + 2)} = x + 1 + \frac{1}{x - 1} - \frac{1}{x + 2} .

Worked example 5: a sum that collapses

Find $S_n = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{n(n + 1)}$ , and evaluate $S_{99}$ .

Solution By the cover-up rule, $\frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1}$ . So

S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n + 1}\right) = 1 - \frac{1}{n + 1} = \frac{n}{n + 1},

because every middle term cancels with the next one (a telescoping sum). Hence $S_{99} = \frac{99}{100}$ .

Dr. Trefor Bazett: Partial fractions with repeated and irreducible quadratic terms (about 10 min). Watch how the form of the decomposition is chosen from the factors before any constants are found.

✦

Practice

Check each answer by putting one value of $x$ into both sides.

$\frac{7x - 1}{(x - 1)(x + 1)}$
$\frac{x + 7}{x^2 - x - 6}$
$\frac{4x^2 + x - 3}{x(x - 1)(x + 3)}$
$\frac{2x + 3}{(x + 1)^2}$
$\frac{x^2 + 1}{x^2(x + 1)}$
$\frac{5x}{(x + 2)(x^2 + 1)}$
$\frac{2x^2 + 3x}{x^2 + x - 2}$ (improper)
Find $\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \cdots + \frac{1}{(2n - 1)(2n + 1)}$ .

Show answersHide answers

$\frac{3}{x - 1} + \frac{4}{x + 1}$ .
$x^{2} - x - 6 = (x - 3) (x + 2)$ : $\frac{2}{x - 3} - \frac{1}{x + 2}$ .
$\frac{1}{x} + \frac{1}{2(x - 1)} + \frac{5}{2(x + 3)}$ ( $A = 1$ , $B = \frac{2}{4}$ , $C = \frac{30}{12}$ ).
$2 x + 3 = 2 (x + 1) + 1$ : $\frac{2}{x + 1} + \frac{1}{(x + 1)^2}$ .
$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1}$ : $B = 1$ , $C = 2$ , $A = - 1$ : $-\frac{1}{x} + \frac{1}{x^2} + \frac{2}{x + 1}$ .
$A = \frac{-10}{5} = -2$ ; $0 = A + B \Rightarrow B = 2$ ; $0 = A + 2C \Rightarrow C = 1$ : $-\frac{2}{x + 2} + \frac{2x + 1}{x^2 + 1}$ .
$2 + \frac{x + 4}{(x - 1)(x + 2)} = 2 + \frac{5}{3(x - 1)} - \frac{2}{3(x + 2)}$ .
$\frac{1}{(2k - 1)(2k + 1)} = \frac{1}{2}\left(\frac{1}{2k - 1} - \frac{1}{2k + 1}\right)$ , so the sum is $\frac{1}{2}\left(1 - \frac{1}{2n + 1}\right) = \frac{n}{2n + 1}$ .

Quick check

Which form is correct for the partial fractions of (x + 4) ÷ ((x − 2)(x² + 3))?

Key points

Make the fraction proper (divide if needed) and factorise the denominator completely.
Distinct linear factor: $\frac{A}{ax + b}$ ; repeated $(a x + b)^{2}$ : $\frac{A}{ax + b} + \frac{B}{(ax + b)^2}$ ; irreducible quadratic: $\frac{Ax + B}{ax^2 + bx + c}$ .
Find constants by substituting roots, then comparing coefficients; check with one value of $x$ .
Partial fractions make sums telescope, and later make rational functions easy to integrate.

1Running addition backwards

Running addition backwards