Recipe

Decompose any rational expression into partial fractions

Use when: the denominator factors into any mix of distinct linear, repeated linear, and irreducible quadratic factors (repeated or not)

Goal: the complete term-by-term decomposition, with every constant solved for, whatever the denominator's factorization looks like

Taught in Irreducible quadratic factors · extends Decompose a rational expression with a repeated linear factor

1. 1

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

   Why: the term-shape argument below only holds for a proper fraction.

2. 2

   Factor q(x) over the reals. For any quadratic factor, compute b^2-4ac before deciding it's irreducible.

   • (x-r), not repeated — one term, \dfrac{A}{x-r}
   • (x-r)^k, repeated — k terms, one per power, constant numerators
   • (ax^2+bx+c) with b^2-4ac<0, not repeated — one term, \dfrac{Bx+C}{ax^2+bx+c}, a LINEAR numerator
   • (ax^2+bx+c)^m with b^2-4ac<0, repeated — m terms, one per power, each with its own linear numerator

   Why: a discriminant of exactly 0 means a repeated REAL root, a linear case in disguise, not a quadratic one. This check has to happen before any term is written down, not after.

3. 3

   Give each irreducible quadratic factor (or each power of one, if it repeats) a numerator of the form Bx+C, never a bare constant.

   Example

   (x-1)(x^2+1) gives \dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}.

   Why: an irreducible quadratic factor's degree-2 denominator needs a degree-1 (linear) numerator to stay proper, the same one-degree-less rule the distinct-linear case used, scaled up.

4. 4

   Cover each distinct linear factor (or the top power of a repeated one) and substitute its root, exactly as in the earlier sections.

   Why: an irreducible quadratic factor has no real root, so cover-up's shortcut is unavailable there. It only ever reaches a linear factor's constant.

5. 5

   Multiply by the full denominator, substitute every constant already known, then match coefficients of like powers of x (or plug in convenient values) for what's left. This is usually the linear-numerator constants sitting over the quadratic factor(s).

   Why: with no real root to substitute at a quadratic factor, matching coefficients is the only method guaranteed to reach a constant living there.

6. 6

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

   Why: catches a wrong numerator shape or a coefficient slip before it costs the whole problem.