Recipe

Decompose a rational expression into partial fractions

Use when: you have a proper rational expression p(x)/q(x) whose denominator factors into distinct linear factors

Goal: the equivalent sum of simple fractions, one term per factor, with every numerator constant pinned down

Taught in Distinct linear factors · a later section extends this recipe with more cases

1. 1

   Compare \deg p to \deg q. Decomposition only applies directly to a proper expression.

   • \deg p < \deg q already — continue straight to Step 2
   • \deg p \ge \deg q — perform polynomial long division first, set the polynomial part aside, and decompose only the proper remainder term

   Why: the whole numerator-shape argument below assumes a proper fraction; skipping this check produces a setup that can never work, no matter what constants you try.

2. 2

   Factor q(x) into its irreducible pieces over the reals.

   • Not yet: a factor that repeats — covered in Repeated linear factors
   • Not yet: an irreducible quadratic factor — covered in Irreducible quadratic factors

   Why: the shape every term needs (a constant numerator here, a whole ladder of terms there) is read directly off this factorization, before a single constant is found.

3. 3

   For each distinct linear factor (x-r), write \dfrac{A}{x-r} with its own unknown constant.

   Example

   (x+2)(x-3) gives \dfrac{A}{x+2} + \dfrac{B}{x-3}.

   Why: a numerator sitting over a degree-1 denominator factor must itself have degree at most 0, a plain constant, or that piece wouldn't be proper on its own.

4. 4

   Cover the factor (x-r) in the ORIGINAL denominator and substitute x=r into what remains; that number is the constant over that factor.

   ⚡ Shortcut

   Coefficient matching (clear every denominator, then match like powers of x) always works too. It's the method the harder cases in the next two sections fall back on whenever cover-up can't reach a constant directly.

   Why: multiplying the identity by (x-r) and evaluating at x=r kills every other term at once, since each one still carries a surviving factor of (x-r) that vanishes there.

5. 5

   Add the finished fractions back over a common denominator and confirm the result matches the original numerator.

   Why: this catches a sign slip or a wrong-root substitution before it costs you the whole problem.