Pure 2.10 - Partial fractions

Pure 2.10 - Partial fractions

One fraction can hide several simpler ones. Learn how to uncover those pieces, handle repeated factors and check that your answer is an identity. This prepares rational functions for later integration, differentiation and series work.

Reverse the common-denominator process

Can you recover two fractions after somebody has added them together? For example,

3x−2+4x+3=3(x+3)+4(x−2)(x−2)(x+3)=7x+1(x−2)(x+3).\frac3{x-2}+\frac4{x+3} =\frac{3(x+3)+4(x-2)}{(x-2)(x+3)} =\frac{7x+1}{(x-2)(x+3)}.

The right-hand side looks more complicated, although its value has not changed. Partial fractions run that calculation backwards.

You already know

To add algebraic fractions, use a common denominator and multiply each numerator by the missing factor. A fraction is undefined wherever its denominator is zero.

The original denominator restrictions survive the rewriting: here x≠2,−3x\ne2,-3. Writing a simpler expression never makes an excluded input legal.

Partial-fraction decomposition

A partial-fraction decomposition expresses a rational function as a sum of fractions with simpler denominators. For example, the two fractions above decompose the single fraction. The pieces form an identity wherever the original expression is defined.

Our denominators contain distinct linear factors or squared linear factors; the decomposition has at most three terms. The original numerator is constant or linear. These are the forms to recognise:

DenominatorForm of the decomposition
(ax+b)(cx+d)(ax+b)(cx+d)Aax+b+Bcx+d\frac A{ax+b}+\frac B{cx+d}
(ax+b)(cx+d)(ex+f)(ax+b)(cx+d)(ex+f)Aax+b+Bcx+d+Cex+f\frac A{ax+b}+\frac B{cx+d}+\frac C{ex+f}
(ax+b)2(ax+b)^2Aax+b+B(ax+b)2\frac A{ax+b}+\frac B{(ax+b)^2}
(ax+b)(cx+d)2(ax+b)(cx+d)^2Aax+b+Bcx+d+C(cx+d)2\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2}

Here each linear factor has non-zero coefficient of xx; factors described as distinct have different roots. The letters AA, BB and CC are constants to find. These decomposition forms are not in the formula booklet: learn them.

A surprising feature is uniqueness. In the opening example, if two decompositions existed, subtracting their cleared identities would give 0=(A−3)(x+3)+(B−4)(x−2)0=(A-3)(x+3)+(B-4)(x-2). Setting x=2x=2 forces A=3A=3; setting x=−3x=-3 forces B=4B=4. There is no second pair waiting to be found.

Key idea

Choose the denominator structure first; the unknown constants come afterwards.

For a repeated factor, the table includes both powers because they provide different numerator contributions when the pieces are recombined.

Use two distinct linear factors

Make the factors work for you: after clearing denominators, their roots switch off unwanted terms. The resulting polynomial identity is valid for every real xx, even though the original fractions have excluded inputs.

Two distinct linear factors

For distinct linear factors and a numerator of lower degree than the denominator,

px+q(ax+b)(cx+d)≡Aax+b+Bcx+d.\frac{px+q}{(ax+b)(cx+d)}\equiv\frac A{ax+b}+\frac B{cx+d}.

The fractions are used only where their denominators are non-zero.

Not in the formula booklet: learn it.

The symbol ≡\equiv means an identity: the expressions agree for every allowed input. Multiplying by the common denominator produces an identity without fractions. We can then choose inputs purely to simplify that identity.

Worked example 1: switch off one term at a time

Question. Express 7x+1(x−2)(x+3)\frac{7x+1}{(x-2)(x+3)} in partial fractions.

Plan. Write one constant over each factor, clear denominators, then substitute the two factor roots.

7x+1(x−2)(x+3)≡Ax−2+Bx+3choose the form7x+1≡A(x+3)+B(x−2)clear denominators15=5Aput x=2−20=−5Bput x=−3\begin{aligned} \frac{7x+1}{(x-2)(x+3)}&\equiv\frac A{x-2}+\frac B{x+3} &&\text{choose the form}\\ 7x+1&\equiv A(x+3)+B(x-2) &&\text{clear denominators}\\ 15&=5A &&\text{put }x=2\\ -20&=-5B &&\text{put }x=-3 \end{aligned}

Why these inputs? Each makes one coefficient term vanish, leaving one unknown.

Answer. A=3A=3, B=4B=4, so

7x+1(x−2)(x+3)≡3x−2+4x+3,x≠2,−3.\frac{7x+1}{(x-2)(x+3)}\equiv\frac3{x-2}+\frac4{x+3},\qquad x\ne2,-3.

Check. The recombined numerator is 3(x+3)+4(x−2)=7x+13(x+3)+4(x-2)=7x+1.

Change just the numerator next, from 7x+17x+1 to 55. The denominator method stays exactly the same, but one constant becomes negative.

Worked example 2: a constant numerator

Question. Express 5(x−2)(x+3)\frac5{(x-2)(x+3)} in partial fractions.

Plan. Use the same denominator form and isolate each constant.

5≡A(x+3)+B(x−2)clear denominators5=5Aput x=25=−5Bput x=−3\begin{aligned} 5&\equiv A(x+3)+B(x-2) &&\text{clear denominators}\\ 5&=5A &&\text{put }x=2\\ 5&=-5B &&\text{put }x=-3 \end{aligned}

Why can the numerator be constant? The coefficients of xx in the two recombined pieces can cancel.

Answer.

5(x−2)(x+3)≡1x−2−1x+3,x≠2,−3.\frac5{(x-2)(x+3)}\equiv\frac1{x-2}-\frac1{x+3},\qquad x\ne2,-3.

Check. (x+3)−(x−2)=5(x+3)-(x-2)=5, so the numerator is correct.

A minus sign belongs to the coefficient, not to a silently changed denominator.

Watch out

The wrong line Bx+3=−Bx+3\frac B{x+3}=\frac{-B}{x+3} changes the function. If a coefficient is negative, keep it: −1x+3-\frac1{x+3} is a valid partial fraction.

Try another constant numerator; keep the denominator factors visible in your working.

Handle up to three distinct factors

Three different roots let you isolate three constants without solving a simultaneous system. A factor such as 2x−32x-3 is just as useful as x−1x-1: solve it carefully before substituting.

Worked example 3: a non-monic factor

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

Plan. Put a constant over each factor and substitute its root into the cleared identity.

5x−11≡A(x+2)(2x−3)+B(x−1)(2x−3)+C(x−1)(x+2)\begin{aligned} 5x-11&\equiv A(x+2)(2x-3)\\ &\quad+B(x-1)(2x-3)+C(x-1)(x+2) \end{aligned}

The products come from cancelling each term's own denominator.

−6=−3Aput x=1−21=21Bput x=−2−72=74Cput x=32\begin{aligned} -6&=-3A &&\text{put }x=1\\ -21&=21B &&\text{put }x=-2\\ -\frac72&=\frac74 C &&\text{put }x=\frac32 \end{aligned}

Why x=32x=\frac32? It is the root of 2x−32x-3, so only the CC term survives.

Answer.

5x−11(x−1)(x+2)(2x−3)≡2x−1−1x+2−22x−3.\frac{5x-11}{(x-1)(x+2)(2x-3)} \equiv\frac2{x-1}-\frac1{x+2}-\frac2{2x-3}.

Check. The combined numerator is

2(x+2)(2x−3)−(x−1)(2x−3)−2(x−1)(x+2)=5x−11.2(x+2)(2x-3)-(x-1)(2x-3)-2(x-1)(x+2)=5x-11.

Change only the numerator to 33. This tests whether you can keep the same structure when the xx term disappears.

Worked example 4: cancellation of quadratic and linear terms

Question. Decompose 3(x−1)(x+2)(2x−3)\frac3{(x-1)(x+2)(2x-3)}.

Plan. Use the same three root substitutions.

3≡A(x+2)(2x−3)+B(x−1)(2x−3)+C(x−1)(x+2)3=−3Aput x=13=21Bput x=−23=74Cput x=32\begin{aligned} 3&\equiv A(x+2)(2x-3)\\ &\quad+B(x-1)(2x-3)+C(x-1)(x+2)\\ 3&=-3A &&\text{put }x=1\\ 3&=21B &&\text{put }x=-2\\ 3&=\frac74 C &&\text{put }x=\frac32 \end{aligned}

Why retain three terms? A simpler numerator does not remove denominator factors.

Answer.

3(x−1)(x+2)(2x−3)≡−1x−1+17(x+2)+127(2x−3).\frac3{(x-1)(x+2)(2x-3)} \equiv-\frac1{x-1}+\frac1{7(x+2)}+\frac{12}{7(2x-3)}.

Check. At x=0x=0, both sides equal 12\frac12; expanding the combined numerator gives the constant 33.

All these identities exclude x=1,−2,32x=1,-2,\frac32. The substitutions were made into polynomials, not the original fractions.

Exam tip

Show the cleared identity before substituting roots. It makes your method visible even if one arithmetic sign slips.

Doubling the constant numerator changes the constants, while the factor roots stay fixed.

Include every power of a repeated factor

A squared factor needs both a first-power and a squared-denominator term. After multiplying out, those terms contribute different polynomials, so one generally cannot replace the other.

A repeated linear factor

For distinct linear factors,

px+q(ax+b)(cx+d)2≡Aax+b+Bcx+d+C(cx+d)2.\frac{px+q}{(ax+b)(cx+d)^2} \equiv\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2}.

Use the identity only where denominators are non-zero.

Not in the formula booklet: learn it.

The repeated root isolates the coefficient over the highest power. Find the remaining coefficient using another input or by comparing coefficients.

Worked example 5: find the missing coefficient

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

Plan. Include all three terms, use the two roots, then substitute zero.

15−3x≡A(x−2)2+B(x+1)(x−2)+C(x+1)18=9Aput x=−19=3Cput x=215=4(2)−2B+3put x=0\begin{aligned} 15-3x&\equiv A(x-2)^2+B(x+1)(x-2)+C(x+1)\\ 18&=9A &&\text{put }x=-1\\ 9&=3C &&\text{put }x=2\\ 15&=4(2)-2B+3 &&\text{put }x=0 \end{aligned}

Why use zero last? Neither root leaves the BB term alive, so a fresh input supplies the missing equation.

Answer. A=2A=2, B=−2B=-2, C=3C=3, giving

15−3x(x+1)(x−2)2≡2x+1−2x−2+3(x−2)2.\frac{15-3x}{(x+1)(x-2)^2} \equiv\frac2{x+1}-\frac2{x-2}+\frac3{(x-2)^2}.

Check. 2(x−2)2−2(x+1)(x−2)+3(x+1)=15−3x2(x-2)^2-2(x+1)(x-2)+3(x+1)=15-3x.

Change only the constant term of the numerator, from 1515 to 66. The same inputs still isolate the same coefficients.

Worked example 6: use coefficient comparison last

Question. Decompose 6−3x(x+1)(x−2)2\frac{6-3x}{(x+1)(x-2)^2}.

Plan. Find AA and CC from roots, then compare the quadratic coefficients.

6−3x≡A(x−2)2+B(x+1)(x−2)+C(x+1)9=9Aput x=−10=3Cput x=20=A+Bcompare x2\begin{aligned} 6-3x&\equiv A(x-2)^2+B(x+1)(x-2)+C(x+1)\\ 9&=9A &&\text{put }x=-1\\ 0&=3C &&\text{put }x=2\\ 0&=A+B &&\text{compare }x^2 \end{aligned}

Why can CC be zero? The general form must allow every power, but the particular function may cancel one of them.

Answer. A=1A=1, B=−1B=-1, C=0C=0, so

6−3x(x+1)(x−2)2≡1x+1−1x−2.\frac{6-3x}{(x+1)(x-2)^2}\equiv\frac1{x+1}-\frac1{x-2}.

Check. The right side is −3(x+1)(x−2)-\frac3{(x+1)(x-2)}, equal to the original after cancelling −3(x−2)-3(x-2). Still exclude x=−1,2x=-1,2.

A zero coefficient is a result of solving the identity, not permission to guess a missing term beforehand.

Watch out

Writing only Ax+1+C(x−2)2\frac A{x+1}+\frac C{(x-2)^2} omits a necessary degree of freedom. Start with Bx−2\frac B{x-2} included; remove it only if the algebra gives B=0B=0.

Now change the numerator's constant once more and solve all three coefficients.

Verify the identity and prepare for later use

A numerical spot-check can find an error, but one successful input does not establish an identity. Recombining the pieces proves equality for every allowed input and catches denominator-sign errors.

Worked example 7: verify by recombining

Question. Verify x+7(x+1)(1−2x)≡2x+1+51−2x\frac{x+7}{(x+1)(1-2x)}\equiv\frac2{x+1}+\frac5{1-2x}.

Plan. Combine the right-hand side and compare its numerator with the left-hand side.

2x+1+51−2x=2(1−2x)+5(x+1)(x+1)(1−2x)common denominator=2−4x+5x+5(x+1)(1−2x)expand=x+7(x+1)(1−2x)collect terms\begin{aligned} \frac2{x+1}+\frac5{1-2x} &=\frac{2(1-2x)+5(x+1)}{(x+1)(1-2x)} &&\text{common denominator}\\ &=\frac{2-4x+5x+5}{(x+1)(1-2x)} &&\text{expand}\\ &=\frac{x+7}{(x+1)(1-2x)} &&\text{collect terms} \end{aligned}

Why recombine? Agreement of the entire numerator proves the identity, rather than testing only selected inputs.

Answer. The identity holds for x≠−1,12x\ne-1,\frac12.

Check. At x=0x=0, each side is 77.

Change only the sign of the second coefficient. That small change gives a different numerator, so a claimed identity must be checked again.

Worked example 8: reject an incorrect identity

Question. Is 2x+1−51−2x\frac2{x+1}-\frac5{1-2x} equal to x+7(x+1)(1−2x)\frac{x+7}{(x+1)(1-2x)}?

Plan. Recombine the proposed fractions without altering their denominator signs.

2x+1−51−2x=2(1−2x)−5(x+1)(x+1)(1−2x)common denominator=−9x−3(x+1)(1−2x)expand and collect\begin{aligned} \frac2{x+1}-\frac5{1-2x} &=\frac{2(1-2x)-5(x+1)}{(x+1)(1-2x)} &&\text{common denominator}\\ &=\frac{-9x-3}{(x+1)(1-2x)} &&\text{expand and collect} \end{aligned}

Why retain the subtraction through the whole bracket? It changes both terms of 5(x+1)5(x+1).

Answer. The claim is false: the numerator is −9x−3-9x-3, not x+7x+7.

Check. At x=0x=0, the proposed pieces total −3-3, whereas the target equals 77.

The opening fraction has now been recovered and checked. Later, splitting rational functions lets you integrate linear-denominator terms as logarithms, differentiate separate negative powers and expand suitable A1−kx\frac A{1-kx} pieces as series. The relevant calculus rules and series validity intervals are taught in their own lessons; the essential preparation here is a correct identity.

Problem-solving

Choose the decomposition to expose the factors, solve the constants in the cleared identity, then recombine before using the result elsewhere.

A different second coefficient gives one final local check.

Quick Recap

  • Use one constant over each distinct linear factor.
  • Include both denominator powers for a squared linear factor.
  • Substitute roots into the cleared polynomial identity, not the original fractions.
  • Recombine to prove your answer, retaining the original excluded inputs.

The opening fraction splits uniquely into 3x−2+4x+3\frac3{x-2}+\frac4{x+3}: clearing denominators reveals the pieces.