Recipe

Decompose a rational expression with a repeated linear factor

Use when: the denominator's factorization includes a linear factor raised to a power greater than 1, like (x-r)^k with k \ge 2

Goal: one numerator term for every power from 1 up to k of the repeated factor, plus the usual single term for every distinct factor, with every constant pinned down

Taught in Repeated linear factors · extends Decompose a rational expression into partial fractions · a later section extends this recipe with more cases

1. 1

   Compare \deg p to \deg q; if the numerator's degree isn't smaller, divide first and decompose only the remainder.

   Why: same precondition as the distinct-factor case: the whole term-shape argument only holds for a proper fraction.

2. 2

   Factor q(x) completely, and for each factor record how many times it repeats.

   • Not yet: an irreducible quadratic factor (repeated or not) — covered in Irreducible quadratic factors

   Why: the repeat count k is exactly how many terms that factor needs. Get this number right before writing a single term.

3. 3

   For a factor (x-r)^k, write \dfrac{A_1}{x-r} + \dfrac{A_2}{(x-r)^2} + \cdots + \dfrac{A_k}{(x-r)^k}: every power from 1 up to k, no skipping the middle rungs. Every distinct (unrepeated) factor still gets its usual single term.

   Example

   (x+1)^2(x-2) gives \dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{C}{x-2}: two rungs for the repeated factor, one term for the distinct one.

   Why: only a full ladder of k independent constants supplies enough freedom to match every numerator clearing denominators can produce; a single term over (x-r)^k can only ever contribute a bare constant.

4. 4

   Cover (x-r)^k (its full power) and substitute x=r to get A_k. Repeat cover-up at each distinct factor's own root.

   Why: only the top rung's covered remainder has nothing left that still vanishes at x=r. Every lower rung's covered remainder still carries a leftover (x-r).

5. 5

   Multiply the whole identity by the full denominator, substitute every constant already found, then plug in one or two convenient x-values (or match coefficients of like powers) for the remaining lower-rung constants.

   ⚡ Shortcut

   Substituting u=x-r turns a lone repeated factor into a bare power u^k; the coefficients of the numerator, rewritten in u, ARE the constants directly, no equations to solve. Only works when nothing else divides the denominator.

   Why: cover-up alone can't reach a lower rung. Whatever it leaves behind still comes from the same identity-matching argument the distinct-factor case used, just with fewer unknowns, since the top rungs are already known.

6. 6

   Add every finished term back over the common denominator and confirm it reproduces the original numerator.

   Why: catches a miscounted rung or a sign slip before it costs the whole problem.