P1.2.10 - Partial fractions

P1.2.10 - Partial fractions

Partial fractions take one algebraic fraction and split it into simpler fractions. This is the reverse of adding fractions over a common denominator. The skill matters because the simpler pieces are much easier to use later in integration, differentiation, and series expansions.

Reverse the common-denominator process

Start with something familiar:

3x−2+4x+3=3(x+3)+4(x−2)(x−2)(x+3)=7x+1(x−2)(x+3).\frac{3}{x-2}+\frac{4}{x+3} =\frac{3(x+3)+4(x-2)}{(x-2)(x+3)} =\frac{7x+1}{(x-2)(x+3)}.

Partial fractions reverse this:

7x+1(x−2)(x+3)=3x−2+4x+3.\frac{7x+1}{(x-2)(x+3)} =\frac{3}{x-2}+\frac{4}{x+3}.

The single rational function has been decomposed into smaller rational functions.

Partial fractions

A partial-fraction decomposition writes one rational function as a sum of simpler fractions whose denominators are factors of the original denominator.

For this Edexcel specification point, the denominator is made from linear factors such as (ax+b), with at most one factor repeated as a square, and the final decomposition has no more than three terms.

Use this pattern table:

Denominator formPartial-fraction form
(ax+b)(cx+d)A/(ax+b) + B/(cx+d)
(ax+b)(cx+d)(ex+f)A/(ax+b) + B/(cx+d) + C/(ex+f)
(ax+b)^2A/(ax+b) + B/(ax+b)^2
(ax+b)(cx+d)^2A/(ax+b) + B/(cx+d) + C/(cx+d)^2

The letters A, B, and C stand for constants to be found. The original numerator in this lesson may be constant or linear, but the partial-fraction numerators over linear factors are constants.

The main method is:

  1. Factorise the denominator if needed.
  2. Write the correct partial-fraction form.
  3. Multiply by the common denominator to get a polynomial identity.
  4. Substitute useful values of x, or compare coefficients.
  5. Write the final expression as partial fractions.

Partial fractions are not a new function. They are the same rational function rewritten in a form that exposes the denominator factors separately.

Use two distinct linear factors

For two different linear factors, write one constant over each factor. Then clear the denominator and choose x-values that make factors zero.

Worked example: express

7x+1(x−2)(x+3)\frac{7x+1}{(x-2)(x+3)}

in partial fractions.

Because the denominator has factors (x-2) and (x+3), write

7x+1(x−2)(x+3)≡Ax−2+Bx+3.\frac{7x+1}{(x-2)(x+3)}\equiv \frac{A}{x-2}+\frac{B}{x+3}.

The symbol \equiv means the two expressions are identical wherever they are defined. Multiply by (x-2)(x+3):

7x+1≡A(x+3)+B(x−2).7x+1 \equiv A(x+3)+B(x-2).

Now choose values that make one term disappear. If x=2, then the B(x-2) term is zero:

15=5A⇒A=3.15=5A \quad \Rightarrow \quad A=3.

If x=-3, then the A(x+3) term is zero:

−20=−5B⇒B=4.-20=-5B \quad \Rightarrow \quad B=4.

Therefore

7x+1(x−2)(x+3)≡3x−2+4x+3.\boxed{\frac{7x+1}{(x-2)(x+3)}\equiv \frac{3}{x-2}+\frac{4}{x+3}}.

You can check by recombining:

3(x+3)+4(x−2)(x−2)(x+3)=7x+1(x−2)(x+3).\frac{3(x+3)+4(x-2)}{(x-2)(x+3)} =\frac{7x+1}{(x-2)(x+3)}.

Use two distinct linear factors Continued

The useful x-values are roots of the denominator factors, but you are not evaluating the original rational function at those roots. The original fraction is undefined there. You are substituting into the identity after multiplying through.

Handle up to three distinct factors

The specification allows denominators with three distinct linear factors, such as

(ax+b)(cx+d)(ex+f).(ax+b)(cx+d)(ex+f).

The form has three terms:

linear or constant numerator(ax+b)(cx+d)(ex+f)≡Aax+b+Bcx+d+Cex+f.\frac{\text{linear or constant numerator}}{(ax+b)(cx+d)(ex+f)} \equiv \frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{ex+f}.

The method is the same, but there are three constants to find. Non-monic factors such as (2x-3) are common, so solve the linear factor carefully.

Worked example: express

5x−11(x−1)(x+2)(2x−3)\frac{5x-11}{(x-1)(x+2)(2x-3)}

in partial fractions.

Set up the form:

5x−11(x−1)(x+2)(2x−3)≡Ax−1+Bx+2+C2x−3.\frac{5x-11}{(x-1)(x+2)(2x-3)} \equiv \frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{2x-3}.

Clear denominators:

5x−11≡A(x+2)(2x−3)+B(x−1)(2x−3)+C(x−1)(x+2).5x-11 \equiv A(x+2)(2x-3)+B(x-1)(2x-3)+C(x-1)(x+2).

Use the roots of the three factors.

When x=1:

−6=A(3)(−1),A=2.-6=A(3)(-1), \qquad A=2.

When x=-2:

−21=B(−3)(−7),B=−1.-21=B(-3)(-7), \qquad B=-1.

When x=3/2, because 2x-3=0:

−72=C(12)(72),C=−2.-\frac{7}{2}=C\left(\frac12\right)\left(\frac72\right), \qquad C=-2.

So

5x−11(x−1)(x+2)(2x−3)≡2x−1−1x+2−22x−3.\boxed{ \frac{5x-11}{(x-1)(x+2)(2x-3)} \equiv \frac{2}{x-1}-\frac{1}{x+2}-\frac{2}{2x-3} }.

The root of (2x-3) is 3/2, not 3. That is one of the easiest places to lose a mark.

Handle up to three distinct factors Continued

Support is starting to fade now: after the form is correct, the remaining work is ordinary algebra. If a convenient root does not give every constant, substitute another simple value such as x=0, or compare coefficients of powers of x.

Include every power of a repeated factor

Repeated linear factors are the main structural trap in this topic. If the denominator contains (cx+d)^2, you need two related terms:

Bcx+d+C(cx+d)2.\frac{B}{cx+d}+\frac{C}{(cx+d)^2}.

Do not write only the squared-denominator term. The first-power term is needed because, after recombining, it contributes a different numerator structure.

Worked example: express

15−3x(x+1)(x−2)2\frac{15-3x}{(x+1)(x-2)^2}

in partial fractions.

The denominator has one single linear factor (x+1) and one repeated linear factor (x-2)^2, so write

15−3x(x+1)(x−2)2≡Ax+1+Bx−2+C(x−2)2.\frac{15-3x}{(x+1)(x-2)^2} \equiv \frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}.

Clear denominators:

15−3x≡A(x−2)2+B(x+1)(x−2)+C(x+1).15-3x \equiv A(x-2)^2+B(x+1)(x-2)+C(x+1).

The roots still help. When x=-1:

18=9A⇒A=2.18=9A \quad \Rightarrow \quad A=2.

When x=2:

9=3C⇒C=3.9=3C \quad \Rightarrow \quad C=3.

There is no remaining root that isolates B, so use a convenient value such as x=0:

15=2(0−2)2+B(0+1)(0−2)+3(0+1).15=2(0-2)^2+B(0+1)(0-2)+3(0+1).

So

15=8−2B+3,15=8-2B+3,

and hence

B=−2.B=-2.

Therefore

15−3x(x+1)(x−2)2≡2x+1−2x−2+3(x−2)2.\boxed{ \frac{15-3x}{(x+1)(x-2)^2} \equiv \frac{2}{x+1}-\frac{2}{x-2}+\frac{3}{(x-2)^2} }.

The repeated factor creates two terms. The root x=2 gives the coefficient of the highest-power term, C/(x-2)^2, but it does not automatically give the coefficient of B/(x-2).

Include every power of a repeated factor Continued

A quick way to check the form before solving is to count the terms. A squared factor contributes two possible denominator powers, so it normally contributes two partial-fraction terms.

Verify the identity and prepare for later use

A partial-fraction answer should be easy to verify. Put the pieces back over the common denominator and check that the numerator matches the original.

Worked example: verify

x+7(x+1)(1−2x)≡2x+1+51−2x.\frac{x+7}{(x+1)(1-2x)} \equiv \frac{2}{x+1}+\frac{5}{1-2x}.

Combine the right-hand side:

2x+1+51−2x=2(1−2x)+5(x+1)(x+1)(1−2x).\frac{2}{x+1}+\frac{5}{1-2x} =\frac{2(1-2x)+5(x+1)}{(x+1)(1-2x)}.

Simplify the numerator:

2(1−2x)+5(x+1)=2−4x+5x+5=x+7.2(1-2x)+5(x+1)=2-4x+5x+5=x+7.

So the decomposition is correct:

2x+1+51−2x≡x+7(x+1)(1−2x).\frac{2}{x+1}+\frac{5}{1-2x} \equiv \frac{x+7}{(x+1)(1-2x)}.

This checking habit matters in exams because an answer with the correct constants but the wrong denominator signs is not a correct decomposition.

Verify the identity and prepare for later use Continued

Partial fractions also prepare expressions for later pure-maths work:

  • In integration, terms such as A/(x+a) lead to logarithms.
  • In differentiation, terms such as A(x+a)^{-1} or B(x+a)^{-2} are easier to differentiate term by term than a single quotient.
  • In series expansions, terms such as A/(1-kx) can be expanded using geometric or binomial forms when the validity condition is satisfied.

Those later applications have their own lessons. For this lesson, the essential skill is to produce and verify the decomposition accurately.

Explain It Back

Use this as a self-explanation check after the section above. It is for diagnosing what you can already explain, not for learning new material from scratch.