1The idea in one paragraph

Adding two simple fractions gives one complicated one: 3/(x − 2) + 2/(x + 3) = (5x + 5)/((x − 2)(x + 3)). Partial fractions is the same calculation run backwards: given the complicated fraction, find the simple ones that add up to it. The guide limits you to the cleanest case, a denominator with at most two different linear factors and a numerator of lower degree. Then the answer always has the form A/(first factor) + B/(second factor), and the whole job is finding the two constants A and B. You do that by writing an identity and choosing values of x that make it easy.

2Adding forwards, splitting backwards

Start with the direction you already know. Add two fractions with linear denominators:

Three things are worth noticing, because they tell you what the reverse process must look like. The denominator of the answer is the product of the two denominators. The numerator of the answer is only linear, one degree less than the quadratic denominator. And once the fractions are combined, the 3 and the 2 have vanished inside 5x + 5. Partial fractions recovers them. Figure 1 shows the two directions.

Why bother? Because the split form is far easier to work with. Figure 2 draws the two pieces and their sum. Each piece is a simple reciprocal graph with one vertical asymptote, and the sum takes one asymptote from each. When you come to integrate in AHL 5.15, 3/(x − 2) and 2/(x + 3) each integrate in one line to a logarithm, while (5x + 5)/((x − 2)(x + 3)) as it stands does not integrate by any rule you know.

3What the guide asks for, exactly

The guide limits partial fractions to fractions that are:

• proper: the degree of the numerator is less than the degree of the denominator. With a quadratic denominator, the numerator is linear (such as 7x − 1) or a constant (such as 5);
• over a denominator with at most two distinct linear factors, such as (x + 1)(x − 3) or (2x − 1)(x + 2).

For such a fraction there are always constants A and B with

(px + q)/((x − a)(x − b)) ≡ A/(x − a) + B/(x − b), where a ≠ b.

The three-bar sign ≡ matters. It says the two sides are equal for every x (except a and b, where neither side is defined). That is a much stronger statement than an equation, and it is what lets you choose convenient values of x in section 4.

Repeated factors such as (x − 1)², quadratic factors that do not factorise such as x² + 1, and improper fractions where the numerator's degree is too high all need different forms. The guide does not ask for any of them, so an HL exam question will stay inside the case above.

Nothing about partial fractions is in the formula booklet.

4Method 1: substitute the roots

Multiply both sides of the identity by the whole denominator. That clears every fraction and leaves an identity between polynomials, which is still true for every x.

Worked example 1. Paper 1. Express (7x − 1)/((x + 1)(x − 3)) in partial fractions.

The trick is in the choice of x, and Figure 3 shows it. Put x = 3 and the bracket (x − 3) becomes zero, so the whole A term disappears and B is left alone. Put x = −1 and the B term disappears. The values to use are always the roots of the denominator, the values that make each factor zero.

Students sometimes object that x = 3 is not allowed, because the original fraction is undefined there. The objection is fair for the fraction but not for the line you substitute into. 7x − 1 ≡ A(x − 3) + B(x + 1) is an identity between two polynomials; it holds for every x, 3 included, because two polynomials that agree at infinitely many points agree everywhere.

Check. Put an easy value, x = 0, into both sides of the answer. Left side: (−1)/((1)(−3)) = ⅓. Right side: 2/1 + 5/(−3) = 2 − 5/3 = ⅓. They agree. The check takes ten seconds and catches most sign errors, so do it every time.

5Method 2: equate coefficients

Expand the right-hand side of the polynomial identity and collect terms.

This works because two polynomials are identical only if their coefficients match term by term. If 7x − 1 and (A + B)x + (−3A + B) are the same expression, the numbers in front of x must be equal, and so must the constants.

Which method to use. Substitution is usually quicker, because each line gives one constant straight away. Equating coefficients is better when a question gives you part of the answer, or asks you to find an unknown in the original fraction, as in Q3 of Try it. Knowing both also gives you a check: find A and B one way and confirm them the other.

6Factorise first, and watch the signs

Most exam questions give the denominator expanded, and the first mark is for factorising it.

Worked example 2. Paper 1. Express 5/(2x² + 3x − 2) in partial fractions.

Check with x = 0: left side 5/(−2) = −2.5; right side 2/(−1) − 1/2 = −2.5.

Two points from this example trip students up. The numerator is a constant, 5, and that is fine: a constant is a polynomial of degree 0, lower than 2. And the root of 2x − 1 is x = ½. Substituting x = 1 or x = 2 would kill neither bracket.

Figure 4 sets out the whole routine. Step 1, checking that the fraction is proper, is quick but not optional: if the numerator were quadratic, the form A/(…) + B/(…) would not work and no values of A and B would satisfy the identity.

Denominators written the other way round. A factor such as (1 − x) or (3 − x) is still linear; its root is x = 1 or x = 3. Keep it in the form the question uses, because the next part of the question is usually built around that form, as the next section shows.

7What partial fractions are for

Almost every partial fractions question has a "hence". Three follow-ups turn up.

Integration (AHL 5.15). ∫ (7x − 1)/((x + 1)(x − 3)) dx becomes ∫ (2/(x + 1) + 5/(x − 3)) dx = 2 ln|x + 1| + 5 ln|x − 3| + c. You will learn the integral of 1/(x + k) in Topic 5; the partial fractions are what make it usable here.

Series expansions (1.10). Each partial fraction is a bracket to the power −1, which the extended binomial theorem expands. Adding two easy expansions is much quicker than expanding the original fraction.

Worked example 3. Paper 1. (a) Express (4 − x)/((1 − x)(1 + 2x)) in partial fractions. (b) Hence find the expansion of (4 − x)/((1 − x)(1 + 2x)) in ascending powers of x, up to and including the term in x³, and state the values of x for which it is valid.

The expansion is valid only where both pieces are valid: |x| < 1 and |x| < ½ together give |x| < ½. Giving |x| < 1, or listing both conditions without combining them, loses the final mark.

Sums of series. A sum such as ∑ 1/(r(r + 1)) looks hard, but 1/(r(r + 1)) = 1/r − 1/(r + 1) (check it by adding). Written that way, the sum collapses, as Figure 5 shows: the −½ from r = 1 cancels the +½ from r = 2, the −⅓ from r = 2 cancels the +⅓ from r = 3, and so on down the list.

A sum that collapses like this is called a telescoping sum. It is not named in the guide, but it is a natural use of partial fractions on Paper 3, and in 1.15 you will prove the same result a second way, by induction.