Resolving into partial fractions

Maths · Partial fractions. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A rational fraction P(x)/Q(x) is proper if the degree of P(x) is less than the degree of Q(x), and improper otherwise. Resolving into partial fractions means writing one fraction as a sum of simpler fractions whose denominators are the factors of Q(x).

The result is an identity: it is true for every value of x, not only for some values as in a conditional equation. This is why you may put in any convenient value of x, or compare coefficients of like powers, to find the constants.

The form you write depends on the factors of the denominator. An improper fraction must be divided out first.

Rules to remember

Proper: degree of numerator < degree of denominator. Improper: degree of numerator ≥ degree of denominator
Improper fraction: divide first to get a polynomial + a proper fraction, then resolve the proper fraction
Non-repeated linear factors: one constant for each factor, A/(x − a) + B/(x − b)
Repeated linear factor (x − a)²: A/(x − a) + B/(x − a)², one term for each power
Irreducible quadratic factor (x² + bx + c): the numerator is linear, (Ax + B)/(x² + bx + c)
To find the constants: multiply both sides by the denominator, then put x equal to the zero of each linear factor or equate coefficients
Identity: true for all values of x. Conditional equation: true only for particular values of x

Common mistake

Students write only one term for a repeated factor, or a constant over a quadratic factor. (x + 2)² needs both B/(x + 2) and C/(x + 2)², and an irreducible quadratic needs Ax + B on top.

Worked example

Resolve (3x + 11)/((x − 3)(x + 2)) into partial fractions.

Let (3x + 11)/((x − 3)(x + 2)) = A/(x − 3) + B/(x + 2), so 3x + 11 = A(x + 2) + B(x − 3).
Put x = 3: 20 = 5A, so A = 4.
Put x = −2: 5 = −5B, so B = −1.
Check: 4(x + 2) − (x − 3) = 3x + 11.

Answer: 4/(x − 3) − 1/(x + 2)

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