Partial Fraction Decomposition Calculator: Breaking Rational Functions Apart

Table of Contents - Partial Fraction Decomposition

How to Use This Calculator
What Is Partial Fraction Decomposition?
How to Perform Partial Fraction Decomposition Manually
Real-World Applications
Scenarios People Actually Run Into
Trade-Offs and Decisions People Underestimate
Common Mistakes and How to Recover
Related Topics
How This Calculator Works
FAQs

How to Use This Calculator - Partial Fraction Decomposition

Enter the numerator polynomial in the first input field using standard notation like "2x + 3" or "x² - 5".

Enter the denominator polynomial in the second input field. Make sure it's factored if the calculator requires factored form, or the calculator will factor it for you.

Click "Calculate" to see the decomposition. The output shows:

Individual partial fractions separated
Constants A, B, C, etc. calculated
The complete decomposition as a sum
Verification that the sum equals the original fraction

For proper fractions only—numerator degree must be less than denominator degree. If improper, perform polynomial division first.

What Is Partial Fraction Decomposition?

Imagine you have a complicated fraction like 5x / [(x + 2)(x - 3)]. Partial fraction decomposition breaks this into simpler pieces: A/(x + 2) + B/(x - 3). It's like reverse addition of fractions.

When you add fractions, you find a common denominator and combine. Decomposition reverses this: you start with the combined form and split it back into separate fractions with simpler denominators.

Why this matters: Integration in calculus becomes much easier with simpler fractions. Inverse Laplace transforms in engineering require this technique. Solving differential equations often needs fractions broken apart.

The key insight is that every rational function—a polynomial divided by a polynomial—can be expressed as a sum of simpler fractions whose denominators are factors of the original denominator.

Understanding partial fractions gives you a powerful tool for simplifying complex rational expressions and solving problems that would otherwise be nearly impossible.

How to Perform Partial Fraction Decomposition Manually

Step 1: Check if the fraction is proper.

A proper fraction has numerator degree less than denominator degree. If not, use polynomial division first.

Example: (x³ + 2) / (x² - 1) is improper. Divide to get x + (x + 2)/(x² - 1), then decompose the remainder fraction.

Step 2: Factor the denominator completely.

Factor into linear factors (x + a) and irreducible quadratic factors (x² + bx + c that can't be factored further).

Example: x³ - x = x(x - 1)(x + 1)

Step 3: Set up the decomposition form.

For each distinct linear factor (x - a), write A/(x - a). For repeated linear factors (x - a)ⁿ, write A₁/(x - a) + A₂/(x - a)² + ... + Aₙ/(x - a)ⁿ. For each irreducible quadratic (x² + bx + c), write (Ax + B)/(x² + bx + c).

Example: 5x / [(x + 2)(x - 3)] becomes A/(x + 2) + B/(x - 3)

Step 4: Multiply both sides by the common denominator.

This eliminates all fractions, giving you a polynomial equation.

Example: 5x = A(x - 3) + B(x + 2)

Step 5: Solve for the constants.

Method A (substitution): Substitute values of x that make factors zero.

When x = 3: 5(3) = A(0) + B(5), so 15 = 5B, thus B = 3 When x = -2: 5(-2) = A(-5) + B(0), so -10 = -5A, thus A = 2

Method B (comparing coefficients): Expand and match coefficients of like powers.

5x = A(x - 3) + B(x + 2) = Ax - 3A + Bx + 2B = (A + B)x + (-3A + 2B)

Coefficients of x: 5 = A + B Constants: 0 = -3A + 2B

Solve this system to get A = 2, B = 3.

Step 6: Write the final decomposition.

5x / [(x + 2)(x - 3)] = 2/(x + 2) + 3/(x - 3)

Example with repeated factors:

(3x + 5) / [(x - 1)²] = A/(x - 1) + B/(x - 1)²

Multiply by (x - 1)²: 3x + 5 = A(x - 1) + B

When x = 1: 3(1) + 5 = A(0) + B, so B = 8

Expand: 3x + 5 = Ax - A + B Coefficient of x: 3 = A Constants: 5 = -A + B, which gives 5 = -3 + 8 = 5 (checks out)

Result: (3x + 5) / [(x - 1)²] = 3/(x - 1) + 8/(x - 1)²

Example with irreducible quadratic:

(2x² + 3x + 4) / [(x + 1)(x² + 1)] = A/(x + 1) + (Bx + C)/(x² + 1)

Multiply by (x + 1)(x² + 1): 2x² + 3x + 4 = A(x² + 1) + (Bx + C)(x + 1)

When x = -1: 2(1) - 3 + 4 = A(2) + 0, so 3 = 2A, thus A = 3/2

Expand right side: A(x² + 1) + (Bx + C)(x + 1) = Ax² + A + Bx² + Bx + Cx + C

Collect: (A + B)x² + (B + C)x + (A + C)

Match coefficients: x²: 2 = A + B, so 2 = 3/2 + B, thus B = 1/2 x: 3 = B + C, so 3 = 1/2 + C, thus C = 5/2 Constant: 4 = A + C = 3/2 + 5/2 = 4 (checks)

Result: (2x² + 3x + 4) / [(x + 1)(x² + 1)] = (3/2)/(x + 1) + (x/2 + 5/2)/(x² + 1)

Real-World Applications

Calculus integration. The integral of 1/[(x - 1)(x + 2)] is difficult directly, but decomposing to A/(x - 1) + B/(x + 2) allows integration as A·ln|x - 1| + B·ln|x + 2| + C.

Laplace transforms. Converting from s-domain back to time-domain in control systems and signal processing requires partial fractions. Transform F(s) = 1/[(s + 1)(s + 3)] into simpler terms for inverse transformation.

Electrical circuit analysis. Transfer functions in frequency domain are rational functions. Decomposition helps analyze circuit behavior at different frequencies.

Differential equations. Solving linear differential equations with the Laplace transform method relies on partial fraction decomposition to convert solutions back to the time domain.

Control theory. System responses, stability analysis, and filter design all use partial fractions to understand how systems behave over time.

Signal processing. Digital filters and frequency analysis break complex transfer functions into manageable pieces using this technique.

Scenarios People Actually Run Into

Integrating a complicated rational function. Your calculus homework has ∫(3x + 7)/[(x + 1)(x - 2)]dx. Direct integration is messy. Decompose first: 2/(x + 1) + 1/(x - 2). Now integrate each term separately.

Converting inverse Laplace transforms. You solved a differential equation and got F(s) = (s + 5)/[s(s + 2)(s + 4)] in s-domain. To get back to time, decompose into A/s + B/(s + 2) + C/(s + 4), then inverse transform each piece.

Simplifying complex rational expressions. You're combining multiple system components and end up with a messy rational function. Decomposing reveals the contribution of each component separately.

Verifying algebra results. Someone gives you 1/(x² - 1) = 1/[2(x - 1)] - 1/[2(x + 1)]. Decompose yourself to verify: x² - 1 = (x - 1)(x + 1), set up A/(x - 1) + B/(x + 1), solve to confirm A = 1/2, B = -1/2.

Analyzing frequency responses. Your electrical circuit has output/input ratio of 10/[(s + 1)(s² + 2s + 2)]. Decompose to understand behavior at different frequencies.

Trade-Offs and Decisions People Underestimate

Factoring the denominator fully. Some people stop at partial factorization. Factor completely into linear and irreducible quadratic factors, or the decomposition will be incomplete.

Choosing substitution versus coefficient matching. Substitution is faster when factors let you choose x values that zero out terms. Coefficient matching is systematic but creates more equations to solve. Use substitution when possible.

Handling improper fractions. If numerator degree is greater than or equal to denominator degree, polynomial division must come first. Skipping this makes the decomposition wrong.

Repeated factor complexity. (x - a)³ requires three terms: A/(x - a) + B/(x - a)² + C/(x - a)³. People sometimes forget to include all powers up to the multiplicity.

Irreducible quadratics. These need numerators of the form Ax + B, not just constants. Missing the linear term makes the system unsolvable.

Common Mistakes and How to Recover

Using wrong numerator for irreducible quadratics. Writing A/(x² + 1) instead of (Ax + B)/(x² + 1). Irreducible quadratics always get linear numerators. Fix: check the degree—linear factors get constants, quadratics get linear numerators.

Not factoring the denominator completely. Decomposing 1/(x² - 4) as A/(x² - 4) doesn't work. Factor to (x - 2)(x + 2) first, then decompose as A/(x - 2) + B/(x + 2).

Forgetting terms for repeated factors. For (x + 1)², writing only A/(x + 1) misses B/(x + 1)². Include all powers from 1 to the multiplicity.

Making arithmetic errors when solving. Substituting x = 2 but calculating wrong values leads to wrong constants. Double-check arithmetic at each substitution.

Setting up improper equation structure. If you write A(x + 2) + B(x - 3) but don't multiply correctly, you'll get wrong coefficients. Expand carefully and verify by plugging values back in.

Skipping the verification step. Always add your partial fractions back together and verify they equal the original. Catches errors immediately.

How This Calculator Works

Input parsing:

Numerator: Parse as polynomial with coefficients
Denominator: Parse and factor into linear and quadratic factors

Factorization algorithm:

Find roots of denominator (real and complex)
Group into linear factors (x - r) for real roots
Group complex conjugate pairs into irreducible quadratics
Identify multiplicities for repeated factors

Setup decomposition form:

For each distinct linear factor: add A/(x - r) term
For repeated linear factor of multiplicity n: add n terms
For each irreducible quadratic: add (Ax + B)/(quadratic) term

Solve for constants:

Multiply equation by common denominator
Use strategic substitution method when possible
Fall back to coefficient matching otherwise
Solve resulting linear system

Output:

Display decomposition with calculated constants
Show verification that sum equals original

FAQs

What is partial fraction decomposition?

Breaking a complicated rational function into a sum of simpler fractions. Like reverse-engineering the process of adding fractions together.

When do I need this technique?

Primarily in calculus for integration, and in engineering for inverse Laplace transforms. Also useful for simplifying complex rational expressions.

What is a proper fraction in this context?

A rational function where the numerator's degree is less than the denominator's degree. Example: x/(x² + 1) is proper, but x²/(x + 1) is not.

What do I do with improper fractions?

Perform polynomial long division first. The quotient stays separate, and you decompose only the remainder fraction.

How do I know what form to use?

Depends on denominator factors. Linear factors get constant numerators. Irreducible quadratics get linear numerators. Repeated factors need multiple terms.

What are irreducible quadratic factors?

Quadratics that can't be factored into real linear factors, like x² + 1 or x² + x + 1. Their discriminant is negative.

Why do repeated factors need multiple terms?

To match all possible partial fraction forms. (x - 2)³ could contribute to each of A/(x - 2), B/(x - 2)², and C/(x - 2)³.

Can I use this for any rational function?

Yes, as long as you can factor the denominator. Some high-degree polynomials are hard to factor by hand, though.

What's the substitution method?

Choosing x values that make certain factors zero, simplifying the equation to solve for one constant at a time. Fast and effective.

What's the coefficient matching method?

Expanding both sides and matching coefficients of like powers of x. Creates a system of equations. More work but always systematic.

How do I verify my answer?

Add the partial fractions back together using common denominator. If you get the original fraction, your decomposition is correct.

What if I get fractions for the constants?

Totally fine. A = 3/2 or B = -5/4 are valid results. Don't round—keep exact fractions.

Can the numerator be higher degree than denominator?

Not for direct decomposition. If it is, it's called improper—do polynomial division first, then decompose the remainder.

What about complex denominators?

If the denominator has complex factors, you can still decompose, but you'll work with complex numbers. Often kept as irreducible quadratics instead.

How do I handle three or more factors?

Same process, more terms. (x + 1)(x - 2)(x + 3) gives A/(x + 1) + B/(x - 2) + C/(x + 3). Set up and solve.

What if two factors are identical?

That's a repeated factor. Handle with multiple terms: (x - 1)² gives A/(x - 1) + B/(x - 1)².

Can I decompose without factoring?

No. The whole technique depends on knowing the factors. Factor first, always.

Why learn this if calculators can do it?

Understanding the process helps you recognize when to use it, verify results, and handle variations that calculators might not support directly.

What's the connection to integration?

Integrals of rational functions often can't be done directly. Partial fractions break them into integrals of 1/(x - a) form, which integrate to logarithms.

Can I decompose multivariable rational functions?

The standard technique works for one variable. Multivariate requires more advanced methods from multivariable calculus.

Additional Notes

Partial fraction decomposition is a reverse-engineering technique. You're undoing the process of adding fractions, which makes seemingly impossible problems manageable.

The technique is algorithmic: factor, set up form, solve for constants, verify. Follow the steps methodically and you'll succeed every time.

Practice factoring first. If you struggle with factoring polynomials, partial fraction decomposition will be harder. Build that foundation skill.

Integration applications make this essential for calculus. Every time you see an integral with a factored denominator, think partial fractions.

In engineering, this technique bridges frequency domain and time domain. Understanding it deeply helps with control systems, signal processing, and circuit analysis.

Always verify your result. Adding the pieces back together takes seconds and confirms you didn't make algebraic errors along the way.

Partial Fraction Decomposition Calculator

📐Types of Partial Fractions

💡Common Examples

💼Applications

🔢Decomposition Steps