Case 2 — g(x) Has a Repeated Linear Factor

The rule. If a linear factor $(ax+b)$ occurs $n$ times in $g(x)$ — that is, $(ax+b)^n$ divides $g(x)$ but $(ax+b)^{n+1}$ does not — then it contributes not one but a whole chain of $n$ partial fractions, with every power from $1$ up to $n$ appearing in a denominator:

where $A_1, A_2, \ldots, A_n$ are constants to be found (with $A_n \neq 0$, otherwise the factor wouldn't really repeat $n$ times). Notice that Case 1 is just the special situation $n=1$ of this same rule.

The method. As before, clear denominators to get a polynomial identity, then solve for the constants. Substituting the repeated root directly gives you the last constant $A_n$ immediately (every other term vanishes), but the remaining constants generally need either equating coefficients, or — often faster — a clean change of variable: if the repeated factor is $(x-a)^n$, substitute $y = x-a$ (so $x = y+a$) everywhere in the numerator. The whole fraction turns into a polynomial in $y$ divided by $y^n$, which splits apart term-by-term into powers of $y$ purely by ordinary division — no simultaneous equations needed at all.

Worked example. Resolve $\dfrac{x^2+1}{(x+2)^3}$ into partial fractions.

Here $(x+2)$ is repeated 3 times, so we expect $\dfrac{A}{x+2} + \dfrac{B}{(x+2)^2} + \dfrac{C}{(x+2)^3}$. Rather than clearing denominators and equating six coefficients, use the substitution trick: let $y = x+2$, so $x = y-2$. Then

so

Substituting back $y = x+2$:

$\frac{x^2+1}{(x+2)^3} = \frac{1}{x+2} - \frac{4}{(x+2)^2} + \frac{5}{(x+2)^3}$ …