Repeated Factors
When a denominator's factor is squared, one term isn't enough to match every numerator — partial fractions needs an extra term, and a different way to pin it down.
Why one term isn't enough
The previous lesson split fractions over distinct linear factors, one constant per factor. When a factor is repeated — squared, in this course — that pattern breaks down. For $(ax+b)(cx+d)^2$, trying $\dfrac{A}{ax+b}+\dfrac{B}{cx+d}$ alone gives only two unknowns, but clearing the denominators produces a numerator that can take three independent values (the coefficients of $x^2$, $x$ and the constant term) — two unknowns can't be tuned to match all three. An extra term, with the repeated factor squared underneath, supplies the missing freedom:
Substituting the repeated root (where $cx+d=0$) still eliminates every term except the one over $(cx+d)^2$, so $C$ is found by direct substitution exactly as before. But only one root exists for that factor, not two — so after finding $C$ this way, and $A$ from the other factor's root, a second technique is needed for $B$: comparing coefficients of the highest power of $x$ (there's rarely a third convenient value of $x$ left to substitute).
A repeated factor with nothing else
When the whole denominator is a single repeated factor, $(ax+b)^2$, the form simplifies to just two terms:
Here both unknowns can be found without any coefficient-comparing at all: clear the denominator to get a linear identity in $A$ and $B$, then either substitute the repeated root (for $B$) and compare the $x$ coefficient (for $A$), or substitute two different values of $x$.
Worked example 1
Worked example 2
With one distinct and one repeated linear factor, there are always exactly two roots available to substitute — one per distinct factor value — no matter how high the power on the repeated factor. Whatever's left over always needs a coefficient comparison.
Worked example 3
A factor like $x^2$ is just as much a repeated linear factor as $(x-3)^2$ — it's $(x-0)^2$. Spot it the same way: does any factor's base appear with a power of 2 (or higher) in the denominator? If so, it needs both a linear and a squared term.
Writing only $\dfrac{A}{ax+b}+\dfrac{B}{cx+d}$ for a denominator with a repeated factor, forgetting the extra $\dfrac{C}{(cx+d)^2}$ term. This leaves too few unknowns to match every possible numerator, and no choice of $A,B$ can make the identity hold for every value of $x$. Always check whether any factor in the denominator is squared before setting up the partial fraction form — and remember that comparing coefficients, not a third substitution, is usually how the last unknown is found.
Check yourself
Express $\dfrac{x+7}{(x+1)^2}$ as partial fractions.
(Hint: substitute x = −1 for one constant, then compare coefficients of x for the other.)