Partial Fractions, Worked From the Cover-Up Method to the Hard Cases

A partial fraction is just a fraction put back into its parts. You already know how to add 1/(x+1) and 1/(x-2) over a common denominator. Partial fractions runs that machine backwards: you are handed the single combined fraction and asked to recover the simpler pieces it came from. It is a setup skill, the thing you do before you integrate or before you expand a binomial series, so a student who is slow here is slow at everything built on top of it.

The reason students find it fiddly is that they try to solve it by brute force, expanding everything and comparing coefficients. There is a faster habit, and it is the same structure-first habit the rest of A-Math rewards: look at the denominator, let its shape tell you the form of the answer, then choose values of x that switch off one unknown at a time.

Start by reading the denominator

The denominator decides the form before you write a single number.

Two distinct linear factors, like (x+1)(x-2), give you one constant over each:

A/(x+1) + B/(x-2).

A repeated linear factor, like (x+1)2, needs both powers:

A/(x+1) + B/(x+1)2.

An irreducible quadratic factor, like (x2+1), takes a linear numerator:

(Ax+B)/(x2+1).

And if the top is the same degree as the bottom or higher, the fraction is improper, so you divide first and split the remainder. Naming the case is half the work. Everything after is bookkeeping.

The cover-up method, on distinct factors

Express (5x + 1) / ((x + 1)(x - 2)) in partial fractions.

Write the form: (5x + 1) / ((x + 1)(x - 2)) = A/(x + 1) + B/(x - 2). Multiply both sides by the denominator to clear it:

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

Now choose values of x that kill one unknown:

Let x = -1. The B term vanishes: 5(-1) + 1 = A(-1 - 2), so -4 = -3A, giving A = 4/3.

Let x = 2. The A term vanishes: 5(2) + 1 = B(2 + 1), so 11 = 3B, giving B = 11/3.

So (5x + 1) / ((x + 1)(x - 2)) = (4/3)/(x + 1) + (11/3)/(x - 2).

A quick check at x = 0: the split gives 4/3 - 11/6 = -1/2, and the original gives 1/((1)(-2)) = -1/2. They agree, so the answer is right.

The repeated factor

Express (4x + 1) / (x + 1)2. The form is A/(x + 1) + B/(x + 1)2, so 4x + 1 = A(x + 1) + B.

Let x = -1: 4(-1) + 1 = B, so B = -3. Comparing the x terms, A = 4.

So (4x + 1) / (x + 1)2 = 4/(x + 1) - 3/(x + 1)2. The repeated factor is the case students forget the second term for, and the marks go with it.

The quadratic factor

Express (2x2 + x + 1) / ((x + 1)(x2 + 1)). The form is A/(x + 1) + (Bx + C)/(x2 + 1). Clearing and substituting x = -1 gives A = 1; comparing coefficients gives B = 1, C = 0.

So (2x2 + x + 1) / ((x + 1)(x2 + 1)) = 1/(x + 1) + x/(x2 + 1). The quadratic factor keeps its linear numerator Bx + C, not a single constant.

The improper case

Express (x2 + 1) / (x2 - 1). Top and bottom are the same degree, so divide first: (x2 + 1)/(x2 - 1) = 1 + 2/(x2 - 1). Then split the proper remainder over (x - 1)(x + 1):

(x2 + 1) / (x2 - 1) = 1 + 1/(x - 1) - 1/(x + 1).

The whole-number 1 out front is the part students drop. Divide before you split, every time the fraction is improper.

The one habit that beats partial fractions

Across all four cases the move is identical: read the denominator, write the matching form, then choose x to switch off unknowns one at a time, only falling back on comparing coefficients for the leftover. Naming the case first is what turns a question that looks long into a question that is short. It is the same reflex GPA trains across the whole of A-Math, where the wrong choice of structure, not the arithmetic, is what usually costs the mark.

If your child can set up the form but stalls on which substitution to make, that is a teachable gap, not a talent ceiling. GPA's Secondary Math programme works on exactly this structure-first reading of a question, with worked A-Math papers behind every lesson. A short diagnostic consult will show you where the gap is.

Worked examples on this page were each checked independently before publishing.