1.5.10 Partial fractions (A-level only)

What you'll learn

How to recognise a rational function and decide whether algebraic division is needed.
How to decompose fractions containing distinct or repeated linear factors.
How to find unknown coefficients by substitution or by comparing coefficients.
How to check that a partial-fraction decomposition is correct.

Prerequisites

Before starting, you should be comfortable with:

factorising quadratic and cubic expressions;
expanding brackets;
solving simultaneous linear equations;
dividing polynomials;
working with identities.

For example,

x^2-x-6=(x-3)(x+2).

Factorising the denominator is usually the first step in partial fractions.

Rational functions and proper fractions

Definition

Rational function

A rational function is a fraction in which both the numerator and denominator are polynomials, such as $3x+1x2−x−6\frac{3x+1}{x^2-x-6}$ .

A rational function is proper if the degree of its numerator is less than the degree of its denominator. The degree of a polynomial is its highest power of $x$ .

For example, $2x+1x2+3x+2\frac{2x+1}{x^2+3x+2}$ is proper because its numerator has degree 1 and its denominator has degree 2.

A rational function is improper if the numerator has degree greater than or equal to the denominator’s degree. You must use polynomial division before decomposing an improper fraction.

Example

Making an improper fraction proper

Express $x2+3x+5x2+x−2\frac{x^2+3x+5}{x^2+x-2}$ as a polynomial plus a proper rational function.

The numerator and denominator both have degree 2, so divide the leading term $x^2$ by $x^2$ . This gives the polynomial term 1.
Subtract the denominator from the numerator:
$x^2+3x+5)-(x^2+x-2)=2x+7.$
Therefore,
$x2+3x+5x2+x−2=1+2x+7x2+x−2.\frac{x^2+3x+5}{x^2+x-2} = 1+\frac{2x+7}{x^2+x-2}.$
The remaining fraction is proper and can now be decomposed after factorising its denominator.

Common Mistake

Decomposing an improper fraction immediately

Do not write down partial fractions until the rational function is proper. If the numerator’s degree is at least the denominator’s degree, divide first.

What are partial fractions?

A complicated rational function can sometimes be written as a sum of simpler rational functions. These simpler fractions are called its partial fractions.

For example, a fraction with denominator $(x - 1) (x + 2)$ may have the form

P(x)(x−1)(x+2)=Ax−1+Bx+2,\frac{P(x)}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2},

where $A$ and $B$ are constants to be found.

Key Idea

The denominator determines the form

Factorise the denominator completely, then include one partial-fraction term for every required factor and power of that factor.

Distinct linear factors

A linear factor has the form $a x + b$ . Distinct linear factors are different from one another.

If the denominator is $(x - a) (x - b)$ , where $a≠ba\neq b$ , use

P(x)(x−a)(x−b)=Ax−a+Bx−b.\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a}+\frac{B}{x-b}.

To find $A$ and $B$ , multiply through by the complete denominator. This creates a polynomial identity.

Definition

Identity

An identity is an equation that is true for every permitted value of the variable. The symbol $≡\equiv$ can be used instead of an equals sign to emphasise this.

Example

Decomposing over two distinct linear factors

Decompose $5x+1(x−1)(x+2)\frac{5x+1}{(x-1)(x+2)}$ into partial fractions.

The denominator contains two distinct linear factors, so write
$5x+1(x−1)(x+2)=Ax−1+Bx+2.\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}.$
Multiply every term by $(x - 1) (x + 2)$ :
$5x+1≡A(x+2)+B(x−1).5x+1\equiv A(x+2)+B(x-1).$
Choose $x = 1$ to eliminate the term containing $B$ :
$6 = 3 A,$
so $A = 2$ .
Choose $x = - 2$ to eliminate the term containing $A$ :
$- 9 = - 3 B,$
so $B = 3$ .
Hence,
$5x+1(x−1)(x+2)=2x−1+3x+2.\frac{5x+1}{(x-1)(x+2)} = \frac{2}{x-1}+\frac{3}{x+2}.$

Tip

Choose roots of the factors

After clearing the denominator, substitute values that make individual brackets zero. This usually finds the constants much faster than expanding everything.

Three distinct linear factors

The same principle works with three factors. If the denominator is $(x - a) (x - b) (x - c)$ , use

P(x)(x−a)(x−b)(x−c)=Ax−a+Bx−b+Cx−c.\frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}.

This is the maximum of three partial-fraction terms required by this specification statement.

Example

Decomposing into three partial fractions

Decompose

2x2+5x−1x(x−1)(x+2).\frac{2x^2+5x-1}{x(x-1)(x+2)}.

The denominator has three distinct linear factors, so set
$2x2+5x−1x(x−1)(x+2)=Ax+Bx−1+Cx+2.\frac{2x^2+5x-1}{x(x-1)(x+2)} = \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}.$
Multiply through by $x (x - 1) (x + 2)$ :
$2x2+5x−1≡A(x−1)(x+2)+Bx(x+2)+Cx(x−1).2x^2+5x-1 \equiv A(x-1)(x+2)+Bx(x+2)+Cx(x-1).$
Substitute $x = 0$ :
$- 1 = - 2 A,$
so $A=12A=\frac12$ .
Substitute $x = 1$ :
$6 = 3 B,$
so $B = 2$ .
Substitute $x = - 2$ :
$- 3 = 6 C,$
so $C=−12C=-\frac12$ .
Therefore,
$2x2+5x−1x(x−1)(x+2)=12x+2x−1−12(x+2).\frac{2x^2+5x-1}{x(x-1)(x+2)} = \frac{1}{2x}+\frac{2}{x-1}-\frac{1}{2(x+2)}.$

Repeated linear factors

A repeated linear factor occurs when the same factor appears more than once, such as $x-2)^2$ .

For every power of the repeated factor, include a separate term:

P(x)(x−a)2=Ax−a+B(x−a)2.\frac{P(x)}{(x-a)^2} = \frac{A}{x-a}+\frac{B}{(x-a)^2}.

If the denominator is $x-a)^2(x-b)$ , the required form is

P(x)(x−a)2(x−b)=Ax−a+B(x−a)2+Cx−b.\frac{P(x)}{(x-a)^2(x-b)} = \frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}.

Common Mistake

Missing the lower power

For a factor $x-a)^2$ , you need terms with both denominators $x - a$ and $x-a)^2$ . Writing only the squared-denominator term gives an incomplete decomposition.

Example

Decomposing with a repeated linear factor

Decompose

3x2−4x+3(x−1)2(x+2).\frac{3x^2-4x+3}{(x-1)^2(x+2)}.

Include both powers of the repeated factor:
$3x2−4x+3(x−1)2(x+2)=Ax−1+B(x−1)2+Cx+2.\frac{3x^2-4x+3}{(x-1)^2(x+2)} = \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}.$
Multiply through by $x-1)^2(x+2)$ :
$3x2−4x+3≡A(x−1)(x+2)+B(x+2)+C(x−1)2.3x^2-4x+3 \equiv A(x-1)(x+2)+B(x+2)+C(x-1)^2.$
Substitute $x = 1$ :
$2 = 3 B,$
so $B=23B=\frac23$ .
Substitute $x = - 2$ :
$23 = 9 C,$
so $C=239C=\frac{23}{9}$ .
Use another convenient value, such as $x = 0$ :
$3 = - 2 A + 2 B + C .$
Substituting $B=23B=\frac23$ and $C=239C=\frac{23}{9}$ gives
$3=−2A+43+239,3=-2A+\frac43+\frac{23}{9},$
so $A=49A=\frac49$ .
Hence,
$3x2−4x+3(x−1)2(x+2)=49(x−1)+23(x−1)2+239(x+2).\frac{3x^2-4x+3}{(x-1)^2(x+2)} = \frac{4}{9(x-1)} +\frac{2}{3(x-1)^2} +\frac{23}{9(x+2)}.$

Comparing coefficients

Substitution is often quickest, but it may not determine every constant immediately. An alternative is to expand the identity and compare the coefficients of equal powers of $x$ .

Example

Finding constants by comparing coefficients

Suppose clearing the denominator gives

4x+7≡A(x+1)+B(x−2).4x+7\equiv A(x+1)+B(x-2).

Expand the right-hand side:
$A (x + 1) + B (x - 2) = (A + B) x + (A - 2 B) .$
Since the identity is true for every $x$ , the coefficients of $x$ must match and the constant terms must match:
$\qquad A-2B=7.$
Subtract the first equation from the second:
$- 3 B = 3,$
so $B = - 1$ . Then $A = 5$ .

Tip

Combine the methods

Use factor roots to find as many constants as possible, then use one convenient value or compare coefficients to find any constant left over.

Checking your decomposition

You can check an answer by recombining the partial fractions over their common denominator. The resulting numerator must match the original numerator exactly.

You can also substitute an allowed value of $x$ into both the original fraction and your decomposition. This is a useful check, but one matching value alone does not prove the identity.

Common Mistake

Excluded values

Values that make the original denominator zero are excluded from the rational function. You may use these values in the cleared polynomial identity to find coefficients, but you cannot substitute them into the original fractions themselves.

Exam technique

In the exam

Check whether the fraction is proper; use polynomial division first if it is not.
Factorise the denominator completely and write the full partial-fraction structure before finding any constants.
Clear the denominator carefully, then substitute roots of the factors wherever possible.
For a repeated factor, include every power up to the highest one shown.
Recombine the fractions or test a permitted value to catch sign and arithmetic errors.

Self review

Check yourself

What partial-fraction form would you use for a denominator of $x+3)^2(x-4)$ ?
Why must an improper rational function be divided before it is decomposed?
How could you find a remaining coefficient if substitution at the factor roots does not determine it?

What you'll learn

How to recognise a rational function and decide whether algebraic division is needed.
How to decompose fractions containing distinct or repeated linear factors.
How to find unknown coefficients by substitution or by comparing coefficients.
How to check that a partial-fraction decomposition is correct.

Prerequisites

Before starting, you should be comfortable with:

factorising quadratic and cubic expressions;
expanding brackets;
solving simultaneous linear equations;
dividing polynomials;
working with identities.

For example,

x^2-x-6=(x-3)(x+2).

Factorising the denominator is usually the first step in partial fractions.

Rational functions and proper fractions

Definition

Rational function

A rational function is a fraction in which both the numerator and denominator are polynomials, such as $3x+1x2−x−6\frac{3x+1}{x^2-x-6}$ .

A rational function is proper if the degree of its numerator is less than the degree of its denominator. The degree of a polynomial is its highest power of $x$ .

For example, $2x+1x2+3x+2\frac{2x+1}{x^2+3x+2}$ is proper because its numerator has degree 1 and its denominator has degree 2.

A rational function is improper if the numerator has degree greater than or equal to the denominator’s degree. You must use polynomial division before decomposing an improper fraction.

Example

Making an improper fraction proper

Express $x2+3x+5x2+x−2\frac{x^2+3x+5}{x^2+x-2}$ as a polynomial plus a proper rational function.

The numerator and denominator both have degree 2, so divide the leading term $x^2$ by $x^2$ . This gives the polynomial term 1.
Subtract the denominator from the numerator:
$x^2+3x+5)-(x^2+x-2)=2x+7.$
Therefore,
$x2+3x+5x2+x−2=1+2x+7x2+x−2.\frac{x^2+3x+5}{x^2+x-2} = 1+\frac{2x+7}{x^2+x-2}.$
The remaining fraction is proper and can now be decomposed after factorising its denominator.

Common Mistake

Decomposing an improper fraction immediately

Do not write down partial fractions until the rational function is proper. If the numerator’s degree is at least the denominator’s degree, divide first.

What are partial fractions?

A complicated rational function can sometimes be written as a sum of simpler rational functions. These simpler fractions are called its partial fractions.

For example, a fraction with denominator $(x - 1) (x + 2)$ may have the form

P(x)(x−1)(x+2)=Ax−1+Bx+2,\frac{P(x)}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2},

where $A$ and $B$ are constants to be found.

Key Idea

The denominator determines the form

Factorise the denominator completely, then include one partial-fraction term for every required factor and power of that factor.

Distinct linear factors

A linear factor has the form $a x + b$ . Distinct linear factors are different from one another.

If the denominator is $(x - a) (x - b)$ , where $a≠ba\neq b$ , use

P(x)(x−a)(x−b)=Ax−a+Bx−b.\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a}+\frac{B}{x-b}.

To find $A$ and $B$ , multiply through by the complete denominator. This creates a polynomial identity.

Definition

Identity

An identity is an equation that is true for every permitted value of the variable. The symbol $≡\equiv$ can be used instead of an equals sign to emphasise this.

Example

Decomposing over two distinct linear factors

Decompose $5x+1(x−1)(x+2)\frac{5x+1}{(x-1)(x+2)}$ into partial fractions.

The denominator contains two distinct linear factors, so write
$5x+1(x−1)(x+2)=Ax−1+Bx+2.\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}.$
Multiply every term by $(x - 1) (x + 2)$ :
$5x+1≡A(x+2)+B(x−1).5x+1\equiv A(x+2)+B(x-1).$
Choose $x = 1$ to eliminate the term containing $B$ :
$6 = 3 A,$
so $A = 2$ .
Choose $x = - 2$ to eliminate the term containing $A$ :
$- 9 = - 3 B,$
so $B = 3$ .
Hence,
$5x+1(x−1)(x+2)=2x−1+3x+2.\frac{5x+1}{(x-1)(x+2)} = \frac{2}{x-1}+\frac{3}{x+2}.$

Tip

Choose roots of the factors

After clearing the denominator, substitute values that make individual brackets zero. This usually finds the constants much faster than expanding everything.

Three distinct linear factors

The same principle works with three factors. If the denominator is $(x - a) (x - b) (x - c)$ , use

P(x)(x−a)(x−b)(x−c)=Ax−a+Bx−b+Cx−c.\frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}.

This is the maximum of three partial-fraction terms required by this specification statement.

Example

Decomposing into three partial fractions

Decompose

2x2+5x−1x(x−1)(x+2).\frac{2x^2+5x-1}{x(x-1)(x+2)}.

The denominator has three distinct linear factors, so set
$2x2+5x−1x(x−1)(x+2)=Ax+Bx−1+Cx+2.\frac{2x^2+5x-1}{x(x-1)(x+2)} = \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}.$
Multiply through by $x (x - 1) (x + 2)$ :
$2x2+5x−1≡A(x−1)(x+2)+Bx(x+2)+Cx(x−1).2x^2+5x-1 \equiv A(x-1)(x+2)+Bx(x+2)+Cx(x-1).$
Substitute $x = 0$ :
$- 1 = - 2 A,$
so $A=12A=\frac12$ .
Substitute $x = 1$ :
$6 = 3 B,$
so $B = 2$ .
Substitute $x = - 2$ :
$- 3 = 6 C,$
so $C=−12C=-\frac12$ .
Therefore,
$2x2+5x−1x(x−1)(x+2)=12x+2x−1−12(x+2).\frac{2x^2+5x-1}{x(x-1)(x+2)} = \frac{1}{2x}+\frac{2}{x-1}-\frac{1}{2(x+2)}.$

Repeated linear factors

A repeated linear factor occurs when the same factor appears more than once, such as $x-2)^2$ .

For every power of the repeated factor, include a separate term:

P(x)(x−a)2=Ax−a+B(x−a)2.\frac{P(x)}{(x-a)^2} = \frac{A}{x-a}+\frac{B}{(x-a)^2}.

If the denominator is $x-a)^2(x-b)$ , the required form is

P(x)(x−a)2(x−b)=Ax−a+B(x−a)2+Cx−b.\frac{P(x)}{(x-a)^2(x-b)} = \frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}.

Common Mistake

Missing the lower power

For a factor $x-a)^2$ , you need terms with both denominators $x - a$ and $x-a)^2$ . Writing only the squared-denominator term gives an incomplete decomposition.

Example

Decomposing with a repeated linear factor

Decompose

3x2−4x+3(x−1)2(x+2).\frac{3x^2-4x+3}{(x-1)^2(x+2)}.

Include both powers of the repeated factor:
$3x2−4x+3(x−1)2(x+2)=Ax−1+B(x−1)2+Cx+2.\frac{3x^2-4x+3}{(x-1)^2(x+2)} = \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}.$
Multiply through by $x-1)^2(x+2)$ :
$3x2−4x+3≡A(x−1)(x+2)+B(x+2)+C(x−1)2.3x^2-4x+3 \equiv A(x-1)(x+2)+B(x+2)+C(x-1)^2.$
Substitute $x = 1$ :
$2 = 3 B,$
so $B=23B=\frac23$ .
Substitute $x = - 2$ :
$23 = 9 C,$
so $C=239C=\frac{23}{9}$ .
Use another convenient value, such as $x = 0$ :
$3 = - 2 A + 2 B + C .$
Substituting $B=23B=\frac23$ and $C=239C=\frac{23}{9}$ gives
$3=−2A+43+239,3=-2A+\frac43+\frac{23}{9},$
so $A=49A=\frac49$ .
Hence,
$3x2−4x+3(x−1)2(x+2)=49(x−1)+23(x−1)2+239(x+2).\frac{3x^2-4x+3}{(x-1)^2(x+2)} = \frac{4}{9(x-1)} +\frac{2}{3(x-1)^2} +\frac{23}{9(x+2)}.$

Comparing coefficients

Substitution is often quickest, but it may not determine every constant immediately. An alternative is to expand the identity and compare the coefficients of equal powers of $x$ .

Example

Finding constants by comparing coefficients

Suppose clearing the denominator gives

4x+7≡A(x+1)+B(x−2).4x+7\equiv A(x+1)+B(x-2).

Expand the right-hand side:
$A (x + 1) + B (x - 2) = (A + B) x + (A - 2 B) .$
Since the identity is true for every $x$ , the coefficients of $x$ must match and the constant terms must match:
$\qquad A-2B=7.$
Subtract the first equation from the second:
$- 3 B = 3,$
so $B = - 1$ . Then $A = 5$ .

Tip

Combine the methods

Use factor roots to find as many constants as possible, then use one convenient value or compare coefficients to find any constant left over.

Checking your decomposition

You can check an answer by recombining the partial fractions over their common denominator. The resulting numerator must match the original numerator exactly.

You can also substitute an allowed value of $x$ into both the original fraction and your decomposition. This is a useful check, but one matching value alone does not prove the identity.

Common Mistake

Excluded values

Exam technique

In the exam

Check whether the fraction is proper; use polynomial division first if it is not.
Factorise the denominator completely and write the full partial-fraction structure before finding any constants.
Clear the denominator carefully, then substitute roots of the factors wherever possible.
For a repeated factor, include every power up to the highest one shown.
Recombine the fractions or test a permitted value to catch sign and arithmetic errors.

Self review

Check yourself

What partial-fraction form would you use for a denominator of $x+3)^2(x-4)$ ?
Why must an improper rational function be divided before it is decomposed?
How could you find a remaining coefficient if substitution at the factor roots does not determine it?

What you'll learn

Prerequisites

Rational functions and proper fractions

What are partial fractions?

Distinct linear factors

Three distinct linear factors

Repeated linear factors

Comparing coefficients

Checking your decomposition

Review 1.5.10 Partial fractions (A-level only) by teaching Genie

1.5.10 Partial fractions (A-level only)

What you'll learn

Prerequisites

Rational functions and proper fractions

What are partial fractions?

Distinct linear factors

Three distinct linear factors

Repeated linear factors

Comparing coefficients

Checking your decomposition

Review 1.5.10 Partial fractions (A-level only) by teaching Genie

1.5.10 Partial fractions (A-level only)

What you'll learn

Prerequisites

Rational functions and proper fractions

What are partial fractions?

Distinct linear factors

Three distinct linear factors

Repeated linear factors

Comparing coefficients

Checking your decomposition

Review 1.5.10 Partial fractions (A-level only) by teaching Genie

1.5.10 Partial fractions (A-level only) Revision Guide

1.5.10 Partial fractions (A-level only)

What you'll learn

Prerequisites

Rational functions and proper fractions

What are partial fractions?

Distinct linear factors

Three distinct linear factors

Repeated linear factors

Comparing coefficients

Checking your decomposition

Review 1.5.10 Partial fractions (A-level only) by teaching Genie

1.5.10 Partial fractions (A-level only) Revision Guide