Cover each factor and evaluate the rest at its root. For (5x − 4) / ((x−1)(x−2)): cover (x−1), put x = 1 into the rest, and get −1. Cover (x−2), put x = 2, and get 6. So the decomposition is −1/(x−1) + 6/(x−2).
Why the cover-up works
It looks like a trick and it is a proof in disguise.
Start from P(x)/((x−1)(x−2)) = A/(x−1) + B/(x−2) and multiply everything by (x−1). The left side loses that factor. The A term loses it too. But the B term still has an (x−1) in the numerator.
Now set x = 1. That (x−1) becomes zero, so the B term vanishes entirely, and what remains is A = P(1)/(1−2).
Covering the factor and substituting is exactly that sequence, done mentally. Every other term is killed because every other term still carries the factor you multiplied by.
The alternative — multiplying out and equating coefficients — gives the same answer and needs a simultaneous system. For three or four factors that is a great deal more work for no more insight.
It has to be proper
Partial fractions only apply to a proper fraction: numerator degree strictly below denominator degree.
If it is not, divide first. Polynomial long division gives a polynomial plus a proper remainder, and the remainder is what gets decomposed.
So (x² + 1)/((x−1)(x−2)) becomes 1 + (3x − 1)/((x−1)(x−2)), and only the second part needs this method.
Skipping the division does not produce a slightly wrong answer — it produces no answer, because the assumed form has too few degrees of freedom to match. This page refuses the input rather than returning something plausible.
Repeated and quadratic factors
Two cases need more terms than this page provides, and it refuses them rather than approximating.
A repeated linear factor (x−r)² needs two terms: A/(x−r) + B/(x−r)². In general a factor repeated k times needs k terms with rising powers, and the cover-up method finds only the highest one directly.
An irreducible quadratic x² + bx + c needs a linear numerator: (Ax + B)/(x² + bx + c). One unknown is not enough, because the factor has two roots and they are complex.
The counting rule is simple: the total number of unknowns always equals the degree of the denominator. If your assumed form has fewer, it cannot work — and that is the check to make before starting.
Why the technique exists
Partial fractions exist to make integration possible, and that is worth stating plainly because it is otherwise a curious piece of algebra with no obvious purpose.
1/((x−1)(x−2)) has no obvious antiderivative. Split into 1/(x−2) − 1/(x−1) it is two logarithms, and the integral falls out immediately.
Every rational function can be integrated this way, which is a genuinely strong result: the antiderivative is always some combination of logarithms, arctangents and rational terms. Nothing else is needed.
The same decomposition serves inverse Laplace transforms, where each simple fraction corresponds to a known transform pair — which is how differential equations get solved in engineering.
Where it gets used
Integration. The standard method for any rational function, and the reason the technique is taught at all.
Laplace transforms. Inverting a transfer function means decomposing it and matching each term to a table entry.
Control theory. A system’s response is read off its poles, and partial fractions separate the contribution of each one.
Series. Telescoping sums often come from a partial fraction split — 1/(n(n+1)) becomes 1/n − 1/(n+1), and the sum collapses.
Sources and methodology
The decomposition is standard algebra; these are the references.
Method. The cover-up method is used rather than equating coefficients, because it needs one substitution per term instead of solving a simultaneous system, and because it makes the reason visible. Improper fractions and repeated roots are refused with their own explanations rather than approximated, since each needs a genuinely different method. The suite verifies every decomposition by recombining it and comparing with the original at sample points away from the poles, which no coefficient error could survive. That engine is verified on every change against 99 hand-written assertions, including that every decomposition recombines to the original rational function at four sample points, and that an improper fraction and a repeated root are each refused with the correct reason. The count and the per-case breakdown are published on the formula verification page.
Read the guide
An improper fraction needs polynomial division first — the Polynomial Division Calculator does that step.