Math
Partial Fraction Decomposition Calculator
Type a numerator and a denominator and the partial fraction decomposition calculator splits the rational function into a sum of simpler fractions.
It factors the denominator, writes one term for each power of each factor (constants over linear factors, Bx + C over irreducible quadratics), solves for the unknowns exactly by comparing coefficients, shows the cover-up shortcut where it applies, and divides first when the fraction is improper.
Any polynomial. If its degree is not smaller than the denominator's, the calculator divides first.
Expanded or factored. Expanded denominators are factored over the rationals; use factored form for degree 5 and above.
Try an example
Result
Partial fractions
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- Setup with unknowns
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- Factored denominator
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- Coefficients
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- Cover-up method
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- Polynomial part (improper only)
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- Factor types
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Student quick launch
Grade planning, algebra checks, and formulas students reach for most.
Study path
Use this calculator with
Follow these when you want the formula behind the answer, a short lesson, or nearby tools in the same topic.
What the partial fraction decomposition calculator solves
Partial fractions reverse the process of adding fractions with different The bottom number of a fraction; it says how many equal parts make one whole.. A proper rational function P(x)/Q(x), with the degree of P smaller than the degree of Q, can be written as a sum of fractions whose denominators are the factors of Q. This form is what you need for integration, inverse Laplace transforms, and telescoping series.
The four cases
| Denominator factor | Terms to write | Example |
|---|---|---|
| Distinct linear (x - a) | A/(x - a) | (x + 5)/((x - 1)(x + 2)) = 2/(x - 1) - 1/(x + 2) |
| Repeated linear (x - a)^n | A₁/(x - a) + A₂/(x - a)^2 + … + Aₙ/(x - a)^n | 1/(x(x + 1)^2) = 1/x - 1/(x + 1) - 1/(x + 1)^2 |
| Irreducible quadratic | (Bx + C)/(x^2 + bx + c) | (3x^2 + 2x + 1)/((x - 1)(x^2 + 1)) = 3/(x - 1) + 2/(x^2 + 1) |
| Improper (deg P ≥ deg Q) | Long division first, then decompose the remainder | x^3/(x - 1)^2 = x + 2 + 3/(x - 1) + 1/(x - 1)^2 |
Finding the unknowns
- Multiply both sides by the full denominator to clear fractions.
- Cover-up method: for a distinct linear factor (x - a), cover it in the original fraction and substitute x = a into what is left. That gives the constant over (x - a) immediately.
- Compare The number multiplying a variable, like the 3 in 3x.: expand the right side and match the coefficient of each power of x with the left side. This gives a linear system, which the calculator solves exactly.
- Substitute the values back into the setup.
How to use it
- Type the The top number of a fraction; it counts how many parts you have. polynomial.
- Type the denominator, either expanded (x^2 + x - 2) or factored ((x - 1)(x + 2)^2). Factored form is required for degree 5 and up, or when the factors have no rational roots.
- Read the decomposition at the top, the term table, the coefficient equations, and the cover-up work. Open the steps for the full derivation.
How to read the answer
Each term in the result has one factor (or a power of it) as its denominator. A coefficient shown as 1/(3(x + 2)) means A = 1/3 over (x + 2). If the original fraction was improper, the polynomial part appears first and the fractions describe only the remainder. Quadratic factors with A number that cannot be written as a fraction of two whole numbers, like π or √2. roots are kept as quadratics; the decomposition is over the rational numbers.
Common mistakes and edge cases
- Writing only A/(x - a)^2 for a repeated factor: you need a term for every power from 1 up to the multiplicity.
- Using a constant over an irreducible quadratic: the numerator must be Bx + C.
- Skipping long division when the numerator's degree is at least the denominator's.
- Factoring x^2 + 1 over the reals: it has no real roots, so it stays quadratic.
- A denominator that does not factor into linear and quadratic pieces over the rationals must be entered in factored form.
Worked examples
Distinct linear factors
(x + 5)/(x^2 + x - 2) = 2/(x - 1) - 1/(x + 2)
Partial fractions
2/(x - 1) - 1/(x + 2)
Repeated linear factor
1/(x(x + 1)^2) = 1/x - 1/(x + 1) - 1/(x + 1)^2
Partial fractions
1/x - 1/(x + 1) - 1/(x + 1)^2
Irreducible quadratic factor
(3x^2 + 2x + 1)/((x - 1)(x^2 + 1)) = 3/(x - 1) + 2/(x^2 + 1)
Partial fractions
3/(x - 1) + 2/(x^2 + 1)
Improper fraction: divide first
x^3/(x - 1)^2 = x + 2 + 3/(x - 1) + 1/(x - 1)^2
Partial fractions
x + 2 + 3/(x - 1) + 1/(x - 1)^2
Fraction coefficients
(2x + 3)/(2x^2 + 5x + 2) = 4/(3(2x + 1)) + 1/(3(x + 2))
Partial fractions
4/(3(2x + 1)) + 1/(3(x + 2))
Linear numerator over a quadratic
(x + 1)/(x^2 + 4) is already a single partial fraction
Partial fractions
(x + 1)/(x^2 + 4)
Denominator that cannot be factored here
x^5 + x + 1 has no rational roots; type it factored instead
Partial fractions
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Frequently asked questions
When do I need long division before partial fractions?+
Whenever the numerator's degree is greater than or equal to the denominator's. Divide to get a polynomial plus a proper remainder fraction, then decompose only the remainder. The calculator does this automatically.
How does the cover-up method work?+
For a distinct linear factor (x - a), cover that factor in the original fraction and substitute x = a into what remains. The number you get is the constant over (x - a). It also gives the highest-power constant for a repeated factor, but not the lower ones.
Why does a repeated factor need several terms?+
Because (x - a)^2 has two 'levels'. A single constant over (x - a)^2 cannot produce every numerator of degree 1, so you need A/(x - a) + B/(x - a)^2.
What goes over an irreducible quadratic?+
A linear numerator Bx + C. A quadratic like x^2 + 1 has no real roots, so it cannot be split into linear factors and needs two unknowns.
Can the calculator factor the denominator for me?+
Yes, over the rational numbers: it finds every rational root and any remaining quadratic. Denominators of degree 5 or more without rational roots must be typed in factored form.
How do I check a partial fraction decomposition?+
Recombine the terms over the common denominator and compare with the original numerator, or substitute a convenient x value into both sides. The steps show the coefficient equations that make the two sides equal.
About this calculator
- Written by
- mathcheck editorial team
- Last reviewed
- September 4, 2026
Method
- Uses the values entered by the user and stable formulas documented on the page.
References
- OpenStax, Precalculus 2e, Section 9.4 “Partial Fractions” — Decomposition rules for linear, repeated linear, and quadratic factors.
- OpenStax, Calculus Volume 2, Section 3.4 “Partial Fractions” — Partial fractions for integration, including improper fractions.
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Last updated: September 4, 2026