Partial Fractions

Engineers who model electrical circuits, control systems and signals constantly meet complicated fractions that are almost unusable as they stand. Their standard first move is to break each one into a sum of simple pieces — which is exactly the technique on this page.

Here is a game. You are handed the fraction

\frac{5x + 1}{(x + 1)(x - 2)}

and told: somebody built this by adding two simple fractions together — one over (x+1), one over (x-2). Take it apart again. That is the whole idea of partial fractions. When you add algebraic fractions you put them over a common denominator and end up with a single, more complicated fraction. Partial fractions run that machine in reverse: they "un-add" the fraction, splitting one complicated rational expression back into a sum of simpler ones.

Why would you ever want to? Because a beast like \frac{5x+1}{(x+1)(x-2)} is horrible to integrate, horrible to expand as a power series, horrible to feed into almost any later machinery — but \frac{A}{x+1} and \frac{B}{x-2} are as easy as fractions get. Split the monster into pieces and every piece becomes a pushover.

The plan: guess the shape, then find the numbers

Suppose the bottom factorises into two distinct linear pieces. Then we can always write

\frac{5x + 1}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}

for some constants A and B. Notice the numerators are just unknown numbers — one over each factor. The whole job is to pin down those two numbers. The method is always the same three moves:

Propose the shape. One unknown constant over each linear factor: \frac{A}{x+1}+\frac{B}{x-2}.
Clear the denominators. Multiply both sides by (x+1)(x-2) to get an identity with no fractions: 5x+1 = A(x-2) + B(x+1).
Find the unknowns. Substitute clever values of x that make a factor vanish (or compare coefficients).

That last step leans on the factor theorem's idea of choosing an x that switches a factor off. When more than one unknown survives at once, you are simply solving by substitution.

Worked example 1 — two distinct linear factors

Decompose \dfrac{5x + 1}{(x + 1)(x - 2)}. Propose the shape and clear the denominators:

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

This is an identity — true for every x — so we may pick whatever value is most convenient. Put x = 2 to switch off the A term:

5(2) + 1 = A(0) + B(3) \;\Rightarrow\; 11 = 3B \;\Rightarrow\; B = \tfrac{11}{3}.

Now put x = -1 to switch off the B term:

5(-1) + 1 = A(-3) + B(0) \;\Rightarrow\; -4 = -3A \;\Rightarrow\; A = \tfrac{4}{3}.

So the split is \dfrac{5x + 1}{(x + 1)(x - 2)} = \dfrac{4/3}{x + 1} + \dfrac{11/3}{x - 2}. Always sanity-check by adding the two pieces back: over the common denominator the top becomes \tfrac{4}{3}(x-2)+\tfrac{11}{3}(x+1) = \tfrac{4x-8+11x+11}{3} = \tfrac{15x+3}{3} = 5x+1. It reassembles into what we started with, so we're right.

See it split

Step through the same decomposition of \frac{5x + 1}{(x + 1)(x - 2)}. We propose the split, clear the denominators, then substitute the two values of x that knock out one unknown at a time.

The cover-up shortcut

For distinct linear factors there is a lightning-fast trick that skips the algebra. To get the constant sitting over (x - 2), put your finger over that factor in the original fraction and evaluate what's left at the value that kills it, x = 2:

B = \left.\frac{5x + 1}{\;(x + 1)\;}\right|_{x = 2} = \frac{5(2)+1}{2+1} = \frac{11}{3}.

Likewise, cover (x + 1) and evaluate the rest at x = -1:

A = \left.\frac{5x + 1}{\;(x - 2)\;}\right|_{x = -1} = \frac{5(-1)+1}{-1-2} = \frac{-4}{-3} = \frac{4}{3}.

Same answers, no simultaneous equations. The cover-up method is really just the substitution trick done in your head — brilliant for distinct linear factors, but (watch out) it only handles them: repeated factors and quadratics still need the full method for their extra terms.

Worked example 2 — a repeated factor

The shape of the decomposition is dictated by the factors on the bottom, and a repeated factor is the classic trap. To split

\frac{3x + 5}{(x - 1)^2}

you must allow both powers of the repeated factor — one over (x-1) and one over (x-1)^2:

\frac{3x + 5}{(x - 1)^2} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2}.

Clear the denominators by multiplying through by (x-1)^2:

3x + 5 = A(x - 1) + B.

Put x = 1 to kill the A term: 3(1)+5 = B, so B = 8. There is no second root to substitute, so compare coefficients of x: the left has 3x, the right has Ax, hence A = 3. The decomposition is

\frac{3x + 5}{(x - 1)^2} = \frac{3}{x - 1} + \frac{8}{(x - 1)^2}.

If you had lazily written only \frac{A}{(x-1)^2}, you'd have thrown away the \frac{3}{x-1} piece entirely and the problem would have had no solution.

Nearly every partial-fractions mistake is a wrong shape chosen before any numbers are found. Match the factor to its term:

Distinct linear factor (x - a) → one constant term \dfrac{A}{x - a}.
Repeated factor (x - a)^2 → you need both \dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}. A cube needs three terms, and so on. One term is not enough.
Irreducible quadratic (x^2 + 1) (one that won't factor) → its numerator must be a linear expression: \dfrac{Bx + C}{x^2 + 1}, not a bare constant.

And one gate before you even start: the numerator's degree must be less than the denominator's. For a top-heavy fraction like \dfrac{x^2 + 1}{(x-1)(x+2)} (degree 2 over degree 2) you must polynomial-divide first to peel off a whole-number part, then decompose the proper remainder. Skip that and every constant you find will be wrong.

Partial fractions look like a fussy algebra exercise, yet they are secretly one of the most useful techniques in all of applied mathematics — because splitting a hard thing into simple pieces is the universal move.

Integration. Nobody can integrate \frac{5x+1}{(x+1)(x-2)} at a glance — but split it and each piece is an easy logarithm, since \int \frac{A}{x-a}\,dx = A\ln|x-a| + C. This is the whole trick behind integrating rational functions.
Differential equations & control theory. Inverting a Laplace transform means decomposing a rational function of s into pieces you recognise — every term becomes an exponential or a sinusoid.
Electrical circuits. The impedance of a circuit is a rational function; partial fractions turn it into a sum of simple building-block responses.

So this "un-adding fractions" trick, learned in a school algebra lesson, quietly runs underneath engineering, physics, and half of a first-year university maths course.

See it explained

Sal Khan works a partial fraction expansion from scratch, finding the unknown constants.