How this instrument works
Partial fraction decomposition rewrites a single rational expression with a factored denominator as a sum of simpler fractions, each with just one factor. For N ⁄ ((x−p)(x−q)), with p and q distinct, the decomposition is A ⁄ (x−p) + B ⁄ (x−q), where A = N ⁄ (p−q) and B = −A. The two coefficients are always exact opposites of each other in this particular two-distinct-factors case, a direct consequence of how the algebra works out.
This rearrangement matters most in calculus, where integrating the original combined fraction directly can be awkward, but integrating each simpler piece separately (each one just a constant over a linear term) is straightforward — partial fractions are the standard first step before integrating a rational function with a factorable denominator.
This page is scoped to the simplest genuinely useful case: a constant numerator and two distinct linear factors in the denominator. Repeated factors (like (x−p)² alone) or three or more factors need a more elaborate setup with additional unknown coefficients, handled by extending this same underlying method rather than this page's direct two-term formula.
- Enter the original expression's constant numerator into the Numerator (constant) field.
- Enter the first root into the First root (p) field.
- Enter the second root into the Second root (q) field.
- Read A and B: the sheet computes both coefficients of the decomposed fractions directly.
Worked example — 1 ⁄ ((x−3)(x−1))
The expression 1 ⁄ ((x−3)(x−1)) decomposes into 0.5 ⁄ (x−3) − 0.5 ⁄ (x−1) — check by recombining: [0.5(x−1) − 0.5(x−3)] ⁄ ((x−3)(x−1)) simplifies in the numerator to 0.5(x−1−x+3) = 0.5×2 = 1, matching the original numerator exactly.
The expression 5 ⁄ ((x−4)(x−2)) decomposes into 2.5 ⁄ (x−4) − 2.5 ⁄ (x−2) — the identical pattern, scaled up by the larger original numerator, 5 instead of 1.
Questions
What is partial fraction decomposition?
A technique for rewriting a single rational expression with a factored denominator as a sum of simpler fractions, each holding just one of the original factors — a rearrangement that makes the expression easier to integrate or analyze term by term.
Why are A and B always exact opposites in this case?
For exactly two distinct linear factors with a constant numerator, the algebra that solves for A and B always produces A = N⁄(p−q) and B = −A — a specific consequence of this particular two-factor setup, not a general rule for every partial fraction problem.
Why does this matter for calculus?
Integrating the original combined fraction directly can be awkward, but integrating each simpler piece separately — a constant divided by a linear term — is straightforward, which is exactly why partial fractions are the standard first step before integrating many rational functions.
What if p and q are equal?
This formula doesn't apply — repeated factors, like (x−p)² alone, need a different decomposition setup with an extra term, since the two-distinct-roots assumption behind this page's formula no longer holds.
Does this handle a numerator that isn't just a constant?
Not directly — this page is scoped to a constant numerator over two distinct linear factors. A numerator that's itself a polynomial (like x, or x+1) needs a more general decomposition setup with additional unknowns to solve for.