For (3x + 5)/((x + 1)(x + 2)), enter 5,3 as the numerator and the factors 1,1;1 and 2,1;1: you get 2/(x + 1) + 1/(x + 2), with x ≠ −1 and x ≠ −2. If what you need is to cancel common factors, use algebraic fractions.
Worked example: split into simple fractions
Split the quotient into simple fractions. Inputs: numerator `5,3`; factors `1,1;1` and `2,1;1`.
- Set up one term per factor. the quotient = A/(x + 1) + B/(x + 2).
- Equate the numerators. 3x + 5 = A(x + 2) + B(x + 1).
- Substitute two values of x. With x = −1 the second term vanishes: 3(−1) + 5 = 2 = A · 1, so A = 2. With x = −2 the first vanishes: 3(−2) + 5 = −1 = B · (−1), so B = 1.
Result: 2/(x + 1) + 1/(x + 2), with x ≠ −1 and x ≠ −2.
Check it with x = 1: the original fraction is 8/6 = 4/3, and 2/2 + 1/3 also gives 4/3. Also, 2(x + 2) + (x + 1) = 3x + 5.
How to enter the data
Coefficients go from lowest to highest degree, separated by commas: first the constant term, then the coefficient of x and, if there is one, that of x². Each factor takes one line in the form coefficients;multiplicity. Coefficients are integers or fractions such as 1/2, never decimals. The limit is total degree 6 in the denominator.
| Expression | Enter it as |
|---|---|
| x | 0,1;1 |
| x − 1 | -1,1;1 |
| x² + 1 | 1,0,1;1 |
| (x + 1)² | 1,1;2 |
| numerator x² + 1 | 1,0,1 |
Note that the constant term comes first: x − 1 is -1,1, not 1,-1. If a factor appears twice, combine it on a single line with its multiplicity, because two identical lines are rejected.
What changes with the denominator
The method is the same in every case: one term for each factor and each power, equate the numerators and solve for the coefficients. What varies is the form of each term.
| Case | Fraction | Decomposition |
|---|---|---|
| Repeated factor | (2x + 1)/(x(x + 1)²) | 1/x − 1/(x + 1) + 1/(x + 1)² |
| Quadratic factor | (x + 2)/(x(x² + 1)) | 2/x + (1 − 2x)/(x² + 1) |
| High-degree numerator | x³/((x + 1)(x + 2)) | x − 3 − 1/(x + 1) + 8/(x + 2) |
Read each row from left to right: the shape of the denominator decides how many terms appear and of what kind. If the denominator shares a factor with the numerator, simplify first with algebraic fractions and decompose afterward. To find the factors of a denominator of degree 3 or higher, the guide on the domain and factoring of algebraic fractions explains how to look for its zeros.
Where to go next
If you need to cancel common factors or work with expressions containing x, use algebraic fractions, which also keeps the restrictions. When there is an equals sign and you are solving for x, go to fraction equations. For a purely numerical expression, with no x, use fraction expressions.
English Wikipedia describes partial fraction decomposition and requires the numerator over an irreducible factor to have degree lower than that factor, so a quadratic factor gets a linear numerator, as in the table above (accessed October 17, 2026).
Frequently asked questions
How do I factor the denominator before using the calculator?
For a quadratic such as x² + 7x + 12, look for two numbers that add to 7 and multiply to 12: they are 3 and 4, so it becomes (x + 3)(x + 4). With degree 3 or higher, try rational roots: divisors of the constant term over divisors of the leading coefficient. Then enter 3,1;1 and 4,1;1.
Does it help with preparing an integral or a Laplace transform?
Yes, it is the step before. With 1/((x − 1)(x + 2)) = (1/3)/(x − 1) − (1/3)/(x + 2), each term is integrated separately and gives (1/3)ln|x − 1| − (1/3)ln|x + 2| + C. This page only decomposes; the integral or the inverse transform is computed separately.
How do I enter a factor whose leading coefficient is not 1, such as 2x + 1?
With its two coefficients, from lowest to highest degree: 1,2;1. For 1/(x(2x + 1)) the result is 1/x − 2/(2x + 1), and 1·(2x + 1) − 2x = 1 gives back the numerator. The factor is shown exactly as you type it, without dividing by 2.
What if the denominator has a factor of degree 3?
The calculator accepts only linear or quadratic factors, so the cubic has to be split first. If it has a rational root, as x³ − 8 does with root 2, it becomes (x − 2)(x² + 2x + 4) and you can enter it. If it has no rational root, it does not fit in the tool.