Home › Calculators › Partial Fraction Decomposition Calculator
Partial Fraction Decomposition Calculator
Break a rational function P(x)/Q(x) into a sum of simple fractions. The calculator factors the denominator, writes the correct form for linear, repeated and irreducible quadratic factors, builds the coefficient equations and solves them exactly.
Last updated:
Examples:
Enter polynomials in x, expanded (x^3 - 3x - 2) or factored ((x+1)^2(x-2)). The denominator is factored over the rationals automatically.
Partial fraction decomposition
Factored denominator
Form to solve
Coefficients
Step by step
Advertisement
Advertisement
How to use the Partial Fraction Decomposition Calculator
Enter the numerator P(x), for example x^2 + 3x + 5.
Enter the denominator Q(x), expanded or factored, for example (x + 1)^2 (x - 2).
Press Decompose (the answer also updates as you type).
Read the decomposition and the coefficient values A, B, C…
Follow the steps from factoring to solving, then copy the answer.
Which form to use for each factor
Factor in the denominator
Terms in the decomposition
Unknowns
(x − a)
A/(x − a)
1
(x − a)³
A/(x − a) + B/(x − a)² + C/(x − a)³
3
(2x + 1)
A/(2x + 1)
1
x² + 4 (irreducible)
(Ax + B)/(x² + 4)
2
(x² + 4)²
(Ax + B)/(x² + 4) + (Cx + D)/(x² + 4)²
4
x³ − 2 (no rational roots)
(Ax² + Bx + C)/(x³ − 2)
3
The total number of unknowns always equals the degree of the denominator, which is why the coefficient system has exactly one solution.
Plug in roots: x = 2 gives 6 = 3A, so A = 2. x = −1 gives −9 = −3B, so B = 3.
Answer: 2/(x − 2) + 3/(x + 1).
Tips for getting it right
Check that the fraction is proper before setting up the form. If it is not, divide first.
Factor completely. A missed factor gives a system with no solution.
Include every power of a repeated factor, not just the highest one.
Mix methods: plugging in roots quickly gives the top-power coefficients of linear factors; matching coefficients finds the rest.
Verify by adding the pieces back over the common denominator. The calculator's check line does this for you.
Frequently asked questions
What is partial fraction decomposition?
It rewrites a rational function P(x)/Q(x) as a sum of simpler fractions whose denominators are the factors of Q(x). For example, (5x − 4)/(x² − x − 2) = 2/(x − 2) + 3/(x + 1). The simpler pieces are much easier to integrate or to transform.
What form do I use for each factor?
A linear factor (ax + b) gets A/(ax + b). A repeated linear factor (ax + b)ⁿ gets one term for each power: A₁/(ax + b) + A₂/(ax + b)² + … + Aₙ/(ax + b)ⁿ. An irreducible quadratic (ax² + bx + c) gets (Bx + C)/(ax² + bx + c), and a repeated quadratic gets one such term for each power.
What if the numerator's degree is too high?
If deg P ≥ deg Q the fraction is improper. Divide first with polynomial long division to get a polynomial plus a proper fraction R(x)/Q(x), then decompose only the proper part. The calculator does this automatically.
How are the coefficients found?
Multiply both sides by the denominator, expand, and set the coefficients of each power of x equal, which gives a system of linear equations. For linear factors you can also plug in each root (the cover-up or Heaviside method), which makes all other terms vanish.
What is an irreducible quadratic factor?
A quadratic like x² + 1 or x² + 2x + 5 whose discriminant b² − 4ac is negative, so it cannot be factored into real linear factors. Its partial fraction needs a linear numerator Bx + C, not just a constant.
Why is partial fraction decomposition used in calculus?
Each simple fraction has a standard antiderivative: A/(x − a) integrates to A·ln|x − a|, A/(x − a)ⁿ to a power of (x − a), and (Bx + C)/(x² + bx + c) to a mix of a logarithm and an arctangent. It is also used for inverse Laplace transforms and for summing telescoping series.
Does it work if the denominator has irrational roots?
Factors are found over the rational numbers. A quadratic with irrational roots such as x² − 2 is kept as a quadratic factor with a (Bx + C) numerator, which is still a correct decomposition over the rationals.
Are all the tools on FreeToolCabin tested for quality and accuracy?
Yes! Every tool on our site undergoes automated Q/A testing in real browser environments to ensure high precision, correct output, and zero errors. If you happen to spot a bug or have a suggestion, please contact us at [email protected] and our team will review it right away.