Partial fractions: decomposition methods and worked examples

Partial fractions reverse the process of adding algebraic fractions. A complicated rational expression is rewritten as a sum of simpler fractions. The value of the expression does not change, but its structure becomes easier to use in algebra, series and integration.

For example,

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

Combining the right-hand side confirms the identity:

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

The aim of decomposition is to discover the numerators $2$ and $3$ systematically.

Before you begin

You should be able to:

• factorise quadratic and simple polynomial expressions;
• add algebraic fractions using a common denominator;
• expand brackets and compare polynomial coefficients;
• use polynomial division when the numerator has equal or greater degree than the denominator.

First decide whether the fraction is proper

A rational expression

$$\frac{P(x)}{Q(x)}$$

is proper if

$$\deg P<\deg Q.$$

It is improper if $\deg P\ge\deg Q$. An improper fraction must be divided first. Partial fraction forms describe the proper remainder, not the original improper expression directly.

For instance,

$$\frac{x^2+3x+5}{x^2+x-2}$$

is improper because both polynomials have degree $2$. By contrast,

$$\frac{3x+5}{x^2+x-2}$$

is proper.

Choosing the correct partial fraction form

Factorise the denominator completely over the real numbers. Its factors determine the form of the decomposition.

Distinct linear factors

For different linear factors,

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

Each linear factor receives a constant numerator.

More generally,

$$\frac{P(x)}{(x-a)(x-b)(x-c)} =\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}.$$

Repeated linear factors

If a factor is repeated, include every power from $1$ up to its multiplicity:

$$\frac{P(x)}{(x-a)^3} =\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{(x-a)^3}.$$

If another factor is present, include its term as well. For example,

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

Skipping an intermediate power produces a form that is not general enough.

Irreducible quadratic factors

If a quadratic factor cannot be factorised into real linear factors, its numerator must be linear:

$$\frac{P(x)}{(x^2+1)(x-2)} =\frac{Ax+B}{x^2+1}+\frac{C}{x-2}.$$

The numerator over a factor must have lower degree than that factor. This gives the useful rule:

• linear denominator factor: constant numerator;
• quadratic denominator factor: linear numerator;
• cubic denominator factor: quadratic numerator, if such a factor is being left irreducible.

At A-level, the most common non-linear case is an irreducible quadratic with a numerator of the form $Ax+B$.

The core decomposition method

For a proper fraction:

1. Factorise the denominator completely.
2. Write the correct partial fraction form.
3. Multiply through by the common denominator.
4. Find the unknown coefficients using convenient substitutions and, where needed, coefficient comparison.
5. Verify the result.

The equation after clearing denominators is a polynomial identity. It is true for every value of $x$, not merely for the values used to find the coefficients.

Distinct linear factors

Worked example 1: two linear factors

Decompose

$$\frac{7x+11}{(x+1)(x+3)}.$$

Step 1: write the correct form.

$$\frac{7x+11}{(x+1)(x+3)} =\frac{A}{x+1}+\frac{B}{x+3}.$$

Step 2: multiply by $(x+1)(x+3)$.

$$7x+11=A(x+3)+B(x+1).$$

Step 3: choose values that eliminate terms.

Set $x=-1$:

$$4=2A,$$

so $A=2$.

Set $x=-3$:

$$-10=-2B,$$

so $B=5$.

Therefore

$$\boxed{\frac{7x+11}{(x+1)(x+3)} =\frac{2}{x+1}+\frac{5}{x+3}}.$$

The substitutions $x=-1$ and $x=-3$ are efficient because each makes one bracket zero.

The cover-up shortcut

For distinct linear factors, the same calculation can be written compactly. To find the coefficient above $x-a$, multiply the original fraction by $x-a$ and then set $x=a$.

For example, if

$$\frac{P(x)}{(x-a)(x-b)} =\frac{A}{x-a}+\frac{B}{x-b},$$

then

$$A=\left.\frac{P(x)}{x-b}\right|_{x=a}.$$

This is often called the cover-up method. It is a shortcut for strategic substitution, not a different theorem. It works immediately for coefficients attached to distinct linear factors. Repeated factors and irreducible quadratics usually require further work.

Worked example 2: three distinct factors

Decompose

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

Write

$$\frac{2x^2-3x-5}{(x-1)(x+2)(x-3)} =\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}.$$

After clearing denominators,

$$2x^2-3x-5 =A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2).$$

Set $x=1$:

$$-6=A(3)(-2),$$

so $A=1$.

Set $x=-2$:

$$9=B(-3)(-5),$$

so $B=\frac35$.

Set $x=3$:

$$4=C(2)(5),$$

so $C=\frac25$.

Hence

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

Fractional coefficients are perfectly acceptable. Do not change a correct answer merely because you expected integers.

Repeated linear factors

Worked example 3: a repeated factor

Decompose

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

Step 1: include both powers of the repeated factor.

$$\frac{3x+5}{(x-1)^2(x+2)} =\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}.$$

Step 2: clear denominators.

$$3x+5=A(x-1)(x+2)+B(x+2)+C(x-1)^2.$$

Step 3: use the roots of the denominator.

Set $x=1$:

$$8=3B,$$

so $B=\frac83$.

Set $x=-2$:

$$-1=9C,$$

so $C=-\frac19$.

These substitutions do not determine $A$, because both $A$ and one other term vanish at each available root.

Step 4: choose one easy additional value.

Set $x=0$ in the identity:

$$5=-2A+2B+C.$$

Substitute $B=8/3$ and $C=-1/9$:

$$5=-2A+\frac{16}{3}-\frac19 =-2A+\frac{47}{9}.$$

Therefore

$$-2A=-\frac29,$$

so $A=\frac19$.

Thus

$$\boxed{ \frac{3x+5}{(x-1)^2(x+2)} =\frac{1}{9(x-1)}+\frac{8}{3(x-1)^2}-\frac{1}{9(x+2)} }.$$

Coefficient comparison as an alternative

Instead of choosing an extra value, expand the identity and compare coefficients of equal powers of $x$. From

$$3x+5=A(x-1)(x+2)+B(x+2)+C(x-1)^2,$$

expansion gives

$$3x+5=(A+C)x^2+(A+B-2C)x+(-2A+2B+C).$$

Since the left side has coefficient $0$ for $x^2$,

$$A+C=0.$$

Together with the other coefficient equations, this determines $A$, $B$ and $C$. Strategic substitution is usually quicker, but coefficient comparison is systematic and is essential when convenient roots do not provide enough information.

Irreducible quadratic factors

A quadratic $ax^2+bx+c$ has no real linear factors when its discriminant is negative:

$$b^2-4ac<0.$$

The numerator above such a quadratic must be linear.

Worked example 4: linear and quadratic factors

Decompose

$$\frac{2x^2+3x+7}{(x-1)(x^2+4)}.$$

Since $x^2+4$ is irreducible over the reals, write

$$\frac{2x^2+3x+7}{(x-1)(x^2+4)} =\frac{A}{x-1}+\frac{Bx+C}{x^2+4}.$$

Clear denominators:

$$2x^2+3x+7=A(x^2+4)+(Bx+C)(x-1).$$

Set $x=1$:

$$12=5A,$$

so $A=\frac{12}{5}$.

Now expand the right side:

$$A x^2+4A+Bx^2-Bx+Cx-C.$$

Collecting coefficients gives

$$(A+B)x^2+(C-B)x+(4A-C).$$

Compare coefficients with $2x^2+3x+7$:

$$A+B=2,$$ $$C-B=3.$$

Using $A=12/5$,

$$B=2-\frac{12}{5}=-\frac25,$$

and then

$$C=3+B=\frac{13}{5}.$$

Therefore

$$\boxed{ \frac{2x^2+3x+7}{(x-1)(x^2+4)} =\frac{12}{5(x-1)}+\frac{-2x+13}{5(x^2+4)} }.$$

The constant equation $4A-C=7$ provides a useful check:

$$4\left(\frac{12}{5}\right)-\frac{13}{5}=7.$$

Improper algebraic fractions

If the numerator’s degree is at least the denominator’s degree, divide first. The result will be

$$\text{polynomial}+\text{proper fraction}.$$

Only the proper fraction is then decomposed.

Worked example 5: divide before decomposing

Decompose

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

First expand the denominator:

$$(x-1)(x+2)=x^2+x-2.$$

The fraction is improper. Divide:

$$2x^2+5x+1=2(x^2+x-2)+(3x+5).$$

Hence

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

Now write

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

Clearing denominators gives

$$3x+5=A(x+2)+B(x-1).$$

Set $x=1$:

$$8=3A,$$

so $A=\frac83$.

Set $x=-2$:

$$-1=-3B,$$

so $B=\frac13$.

The complete decomposition is

$$\boxed{ \frac{2x^2+5x+1}{(x-1)(x+2)} =2+\frac{8}{3(x-1)}+\frac{1}{3(x+2)} }.$$

The polynomial term $2$ is part of the answer. Omitting it changes the function.

When the denominator is not already factorised

Factorisation is part of the method, not an optional preliminary.

Worked example 6: expose the linear factors

Decompose

$$\frac{4x+1}{x^2-x-6}.$$

Factorise:

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

Then

$$\frac{4x+1}{(x-3)(x+2)} =\frac{A}{x-3}+\frac{B}{x+2}.$$

Clear denominators:

$$4x+1=A(x+2)+B(x-3).$$

Set $x=3$:

$$13=5A,$$

so $A=13/5$.

Set $x=-2$:

$$-7=-5B,$$

so $B=7/5$.

Therefore

$$\boxed{ \frac{4x+1}{x^2-x-6} =\frac{13}{5(x-3)}+\frac{7}{5(x+2)} }.$$

Efficient ways to find coefficients

Several methods are valid. Choose the shortest reliable combination.

Strategic substitution

After clearing denominators, substitute roots of the denominator first. Each such value usually eliminates several terms. This is the fastest method for distinct linear factors.

Compare coefficients

Expand both sides and equate coefficients of $x^n$. This always works when the partial fraction form is correct. It is especially useful for irreducible quadratic factors and repeated factors.

Substitute an easy extra value

After roots have provided most coefficients, $x=0$, $x=1$ or another simple value can be quicker than a full expansion.

Use leading coefficients

Sometimes the coefficient of the highest power gives an immediate short equation. For example, in

$$P(x)=A(x+1)(x+2)+B(x-3)(x+2)+C(x-3)(x+1),$$

the coefficient of $x^2$ on the right is $A+B+C$. Comparing it with the leading coefficient of $P(x)$ can supply a quick check or a missing equation.

A strong solution often mixes methods: use roots first, then one coefficient or one easy substitution.

How to verify a decomposition

You should be able to verify your answer in at least two ways.

Recombine the fractions

Put the partial fractions over the original common denominator and simplify the numerator. It should match exactly.

Check a permitted numerical value

Choose a value not excluded from the domain and evaluate both forms. This detects many arithmetic errors, although one matching value does not prove an identity.

Check structure

Confirm that:

• every denominator factor is represented;
• every repeated power is present;
• each numerator has lower degree than its denominator factor;
• an improper original fraction includes its polynomial quotient.

Worked example 7: verify without full expansion

Suppose you obtain

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

Recombine the right side:

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

The numerator matches, so the decomposition is correct for all $x$ in the common domain, namely $x\ne2,-1$.

Why partial fractions help with integration

Partial fractions turn rational functions into standard integrable forms. For example,

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

leads to

$$\int\frac{5x+1}{(x-1)(x+2)}\,dx =2\ln|x-1|+3\ln|x+2|+C.$$

Repeated factors produce negative powers, while irreducible quadratics may require logarithmic, inverse tangent or further algebraic techniques. The full integration strategy is developed in integration using partial fractions.

Common misconceptions

”Every numerator is a constant”

Only a linear denominator factor has a constant numerator. An irreducible quadratic factor needs a general linear numerator $Ax+B$.

”A repeated factor appears only once”

For $(x-a)^3$, include terms over $(x-a)$, $(x-a)^2$ and $(x-a)^3$. Missing powers make the decomposition incomplete.

”Cover-up solves every coefficient”

Cover-up is immediate for distinct linear factors and can isolate the coefficient over the highest power of a repeated factor. It does not by itself determine all coefficients in every repeated or quadratic case.

”I can decompose an improper fraction directly”

Divide first. Otherwise the proposed right side tends to zero as $|x|$ becomes large while the original fraction may tend to a non-zero constant or grow without bound.

”The roots used for substitution are forbidden, so I cannot use them”

Those values are forbidden in the original rational expression, but after multiplying through you obtain a polynomial identity. Substituting them into that identity is valid and deliberately eliminates terms.

”Matching one numerical value proves the answer”

One value is only a check. A valid derivation comes from the polynomial identity, coefficient comparison or equivalent algebra.

Check your understanding

Question 1

Decompose

$$\frac{9x-1}{(x-2)(x+3)}.$$

Question 2

Write down the correct partial fraction form, without finding coefficients, for

$$\frac{x^2+1}{(x-4)^3(x^2+9)}.$$

Question 3

Decompose

$$\frac{x+7}{x(x+1)^2}.$$

Question 4

Decompose fully

$$\frac{x^2+4x+7}{x^2+3x+2}.$$

Question 5

Find $A$, $B$ and $C$ if

$$\frac{x^2+2x+5}{(x+1)(x^2+1)} =\frac{A}{x+1}+\frac{Bx+C}{x^2+1}.$$

Answers

1. Write

$$\frac{9x-1}{(x-2)(x+3)} =\frac{A}{x-2}+\frac{B}{x+3}.$$

Then

$$9x-1=A(x+3)+B(x-2).$$

Setting $x=2$ gives $17=5A$, so $A=17/5$. Setting $x=-3$ gives $-28=-5B$, so $B=28/5$. Therefore

$$\frac{9x-1}{(x-2)(x+3)} =\frac{17}{5(x-2)}+\frac{28}{5(x+3)}.$$

2. The complete form is

$$\frac{A}{x-4} +\frac{B}{(x-4)^2} +\frac{C}{(x-4)^3} +\frac{Dx+E}{x^2+9}.$$

3. Write

$$\frac{x+7}{x(x+1)^2} =\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}.$$

Clearing denominators gives

$$x+7=A(x+1)^2+Bx(x+1)+Cx.$$

Setting $x=0$ gives $A=7$. Setting $x=-1$ gives $C=-6$. Comparing coefficients of $x^2$ gives $A+B=0$, so $B=-7$. Hence

$$\frac{x+7}{x(x+1)^2} =\frac7x-\frac7{x+1}-\frac6{(x+1)^2}.$$

4. Divide first:

$$\frac{x^2+4x+7}{x^2+3x+2} =1+\frac{x+5}{(x+1)(x+2)}.$$

Now

$$\frac{x+5}{(x+1)(x+2)} =\frac4{x+1}-\frac3{x+2}.$$

Therefore

$$\frac{x^2+4x+7}{x^2+3x+2} =1+\frac4{x+1}-\frac3{x+2}.$$

5. Clearing denominators gives

$$x^2+2x+5=A(x^2+1)+(Bx+C)(x+1).$$

Set $x=-1$ to obtain $4=2A$, so $A=2$. Comparing coefficients gives

$$A+B=1,\qquad B+C=2,\qquad A+C=5.$$

Thus $B=-1$ and $C=3$.

Final checklist

Before finalising a decomposition, check that you have:

• classified the fraction as proper or improper;
• divided first if necessary;
• factorised the denominator completely;
• included every distinct factor and every power of a repeated factor;
• used a numerator whose degree is one less than its denominator factor;
• cleared denominators correctly;
• used strategic substitutions before doing unnecessary expansion;
• verified the completed identity.

What to learn next

• Strengthen division and factorisation in polynomials and algebraic division.
• Use decompositions to integrate rational functions in integration using partial fractions.
• Meet partial fractions in series work through rational binomial expansion.