Partial Fraction Decomposition Calculator
Enter the coefficients of your rational function's numerator and denominator, choose one of the four standard decomposition forms, and get the constants A, B, C (and D) instantly with a full step-by-step worked solution. Covers distinct linear factors, repeated linear factors, linear plus irreducible quadratic, and three distinct linear factors.
Formula
Worked example
For (5x+1)/[(x+1)(x-2)]: substitute x=-1 to get A = (5(-1)+1)/(-1-2) = -4/-3 = 4/3, then x=2 for B = (5(2)+1)/(2+1) = 11/3. Result: (4/3)/(x+1) + (11/3)/(x-2).
What is partial fraction decomposition?
Partial fraction decomposition is an algebraic technique that rewrites a rational function, one polynomial divided by another, as a sum of simpler fractions. Each simpler fraction has a factor of the original denominator as its own denominator and a numerator that is either a constant (for linear factors) or a linear polynomial (for irreducible quadratic factors). The idea mirrors reversing the process of adding fractions: instead of finding a common denominator to combine fractions, you go the other direction and split a complex fraction apart. The technique appears in calculus (integrating rational functions), differential equations (solving with Laplace transforms), signal processing (analyzing transfer functions), and control engineering (inverse Laplace transforms of system responses).
How to perform partial fraction decomposition: a step-by-step guide
Step 1: Make sure the rational function is proper, meaning the degree of the numerator is less than the degree of the denominator. If it is not, do polynomial long division first to extract a polynomial part, then decompose only the remainder. Step 2: Factor the denominator completely into linear factors (ax+b) and irreducible quadratic factors (ax^2+bx+c where b^2-4ac < 0). Step 3: Write the partial fraction template. Each distinct linear factor (x+a) contributes a term A/(x+a). A repeated linear factor (x+a)^n contributes n terms: A/(x+a) + B/(x+a)^2 + ... + N/(x+a)^n. Each irreducible quadratic factor contributes a term (Ax+B)/(quadratic). Step 4: Multiply both sides by the full denominator to clear all fractions, giving a polynomial identity. Step 5: Solve for the unknown constants. You can substitute the roots of each linear factor into the identity (the cover-up method), match coefficients of each power of x, or use a combination of both.
The cover-up method for linear factors
For a denominator with distinct linear factors, the cover-up method is the most efficient approach. To find the constant A for the factor (x+a), mentally cover that factor in the original denominator and substitute x = -a everywhere else. The value of the resulting expression is A. For example, to decompose (5x+1)/[(x+1)(x-2)]: cover (x+1) and substitute x=-1: (5(-1)+1)/((-1)-2) = -4/-3 = 4/3 = A. Then cover (x-2) and substitute x=2: (5(2)+1)/((2)+1) = 11/3 = B. The answer is (4/3)/(x+1) + (11/3)/(x-2). This technique works because substituting the root of a factor makes all other terms vanish, leaving only the one constant you want.
Irreducible quadratic factors and why they need Cx+D
An irreducible quadratic factor ax^2+bx+c has no real roots (its discriminant b^2-4ac is negative). Because it cannot be broken into real linear factors, its contribution to the decomposition must have a numerator of degree one less than the quadratic, that is, a linear numerator Cx+D rather than a plain constant. The constants C and D are found by matching coefficients of each power of x in the polynomial identity you get after clearing denominators. If you mistakenly used just a constant D in the numerator, the left and right sides of the identity would have different numbers of unknowns, and the system would be inconsistent or underdetermined.
Partial fraction templates by denominator type
| Denominator form | Partial fraction template | Unknowns |
|---|---|---|
| (x+a)(x+b), a ≠ b | A/(x+a) + B/(x+b) | A, B |
| (x+a)^2 | A/(x+a) + B/(x+a)^2 | A, B |
| (x+a)(x^2+bx+c), irreducible quad | A/(x+a) + (Cx+D)/(x^2+bx+c) | A, C, D |
| (x+a)^3 | A/(x+a) + B/(x+a)^2 + C/(x+a)^3 | A, B, C |
| (x+a)(x+b)(x+c), distinct | A/(x+a) + B/(x+b) + C/(x+c) | A, B, C |
| (x^2+bx+c)^2, repeated quadratic | (Ax+B)/(x^2+bx+c) + (Cx+D)/(x^2+bx+c)^2 | A, B, C, D |
Standard forms used in partial fraction decomposition. Choose the template that matches your factored denominator.
Frequently asked questions
What if the degree of the numerator is greater than or equal to the degree of the denominator?
Partial fraction decomposition only applies directly to proper rational functions, where the numerator degree is strictly less than the denominator degree. If the numerator degree is equal to or greater, first perform polynomial long division to write the expression as a polynomial plus a proper rational remainder, then apply partial fractions only to the remainder part.
How do I check my partial fraction decomposition is correct?
Combine your partial fractions back over a common denominator and simplify the numerator. If you get the original numerator, the decomposition is correct. Alternatively, substitute a few numerical values of x into both the original expression and the decomposed form and confirm they match.
What is the cover-up method and when does it work?
The cover-up method (Heaviside cover-up) is a shortcut for finding constants of distinct linear factors. You "cover" each factor in the denominator, substitute the root of that factor into the remaining expression, and the result is the constant for that term. It only works directly for simple (non-repeated) linear factors. For repeated factors or quadratic factors you still need to match coefficients.
Why does a repeated factor (x+a)^2 give two terms instead of one?
A repeated factor of multiplicity n generates n separate partial fractions: A/(x+a), B/(x+a)^2, and so on up to N/(x+a)^n. This is because the full general form for a term over (x+a)^n must allow every power from 1 to n in the denominator to be able to represent any numerator polynomial of degree less than n. Omitting any power would leave you unable to match all coefficients in the polynomial identity.
Where is partial fraction decomposition used in practice?
The most common use in calculus is integrating rational functions: each simple term A/(x+a) integrates to A*ln|x+a|, and each quadratic term integrates to an arctangent or a combination of logarithm and arctangent. In engineering, partial fractions are central to the inverse Laplace transform, which converts transfer functions in the frequency domain back into time-domain signals. Control engineers and circuit analysts use it constantly.
Can this calculator handle complex or imaginary roots?
This calculator works over the real numbers and is designed for the four most common forms: two or three distinct real linear factors, a repeated real linear factor, and a linear factor combined with an irreducible real quadratic. Expressions requiring complex roots are handled indirectly through the irreducible quadratic form, which keeps the arithmetic real.