Partial Fractions
One tangled fraction, several unwanted denominators — split it apart into simple pieces, and a whole family of later problems suddenly gets easy.
Splitting a fraction apart
Adding $\dfrac{2}{x-1}$ and $\dfrac{3}{x+2}$ over a common denominator gives a single, more complicated fraction. Partial fractions reverses this: starting from a combined fraction with a factorised denominator, it's rewritten as a sum of simpler fractions, each with one linear factor underneath. This "unmixing" is one of the most useful algebraic skills in the course — later needed for integrating rational functions and binomially expanding them.
Distinct linear factors
For a denominator that is a product of distinct linear factors, $(ax+b)(cx+d)(ex+f)$, the decomposition takes the form
for constants $A$ and $B$ to be found (with a third term added for a third factor). The unknowns are found by multiplying both sides by the original denominator to clear all fractions, then either substituting convenient values of $x$ (each root of a factor eliminates all but one unknown) or comparing coefficients of matching powers of $x$.
Substituting $x=$ (the root of a factor) makes that factor's own term vanish on the right-hand side, and every other term on the right also vanishes since it's still multiplied by that now-zero factor. This isolates a single unknown constant instantly — far quicker than expanding everything and comparing coefficients, though both methods always agree.
Every case in this lesson has a numerator that is a constant or linear expression, one degree lower than the total degree of the denominator ("proper" fractions) — so the general form above always applies directly, without needing an extra polynomial term out front. If a numerator's degree ever matched or exceeded the denominator's, algebraic division would be needed first to pull out a polynomial part before what's left could be split into partial fractions — the next two lessons cover exactly that, and a repeated factor in the denominator.
Worked example 1
Choosing $x$ equal to each factor's root in turn is not a coincidence — it's a deliberate strategy that eliminates all but one unknown per substitution. Always identify these two roots ($x=2$ and $x=-3$ here) before doing any other algebra.
Worked example 2
With three or more unknowns, it's always possible to find each one directly by substitution — there's never a need to compare coefficients, provided the denominator factorises into distinct linear factors and each root is used in turn. Comparing coefficients becomes essential once a repeated factor is involved, as Example 3 shows.
Once a rational function is written as separate simple fractions, each piece becomes far easier to work with individually — $\displaystyle\int\dfrac{A}{ax+b}\,dx$ is a standard log integral, and $\dfrac{A}{ax+b}$ has a clean binomial expansion, whereas the original combined fraction resists both directly. This is precisely why partial fractions appear as a stepping-stone technique before both topics later in the course.
Setting up the wrong number of unknowns for the number of factors — for $n$ distinct linear factors in the denominator, there must be exactly $n$ constants, one per factor. Missing one out, or adding an extra term that isn't justified by the denominator, leaves an identity that can't be solved consistently for every value of $x$. Always count the factors first, and check each one is genuinely linear and distinct before writing down the partial fraction form.
Check yourself
Express $\dfrac{4x-1}{(x-1)(x+1)}$ as partial fractions.
(Hint: substitute x = 1 and x = −1 in turn.)