Repeated Factors

Calculus 2 · Axiom Academy

Why a factor like (x − 1)² needs more than one term — and how to find them. Take a concrete example with a repeated factor in the denominator: It is tempting to treat the repeated factor like a distinct factor and use a single term: Clearing the denominator turns this into x+3 = A . But A is a single constant, and the left side is x+3 — it changes with x . Watch a constant try to keep up with a line that climbs. 2. The Fix: One Term per Power The key idea is to include a term for every power of the repeated factor, from 1 up to its multiplicity. For (x-1)^2 that means two terms: The denominator (x-1)^2 unstacks into one rung for each power, and each rung gets its own unknown constant. This pattern holds for any multiplicity. A factor (x-a)^k contributes k terms — one for each power from 1 to k : As the multiplicity k grows, the ladder of terms grows with it — never fewer terms than the power demands. For example, has multiplicity 3 , so it needs three terms — and they work out to . 4. Working Through the Example Back to . Multiply both sides by (x-1)^2 : Expand the right side and group by powers of x : Now match coefficients on each side. The animation lines up the x -channel and the constant channel until both agree. The reason both terms are required comes from how they recombine over the common denominator (x-1)^2 . Each contributes to the numerator in a different way: contributes A(x-1) — a term that carries the x -part. contributes B — a flat constant, independent of x .

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