What does this calculator decompose?

This calculator decomposes a proper rational function into a sum of partial fractions. It shows the decomposition template, solves the unknown coefficients, and then verifies the result numerically at sample x-values away from denominator roots.

• It supports distinct linear factors such as (x-1)(x+2).
• It supports repeated linear factors such as (x-1)^2 or (x+3)^3.
• It supports irreducible quadratic factors such as x²+1 or x²+2x+5.
• It does not auto-factor an arbitrary denominator typed as one free-text polynomial.
• It rejects improper fractions and points the user toward polynomial division first.

What is partial fraction decomposition?

Partial fraction decomposition rewrites a rational function as a sum of simpler rational terms. It is commonly used in integration, inverse Laplace transforms, algebraic manipulation, and rational-function analysis.

A single fraction with a factored denominator may be hard to work with. After decomposition, each part has a simpler denominator and a predictable numerator form.

Why does the denominator use factor cards?

The denominator uses factor cards because automatic factorization of any user-typed polynomial is outside the safe scope of this calculator. The user supplies the factor structure, and the calculator builds the partial fraction system from that structure.

This is a deliberate product choice. It lets the calculator cover the core classroom and competitor cases while avoiding a fragile parser or CAS promise that could fail on non-nice polynomials.

• A linear factor card represents a factor of the form x-r.
• An irreducible quadratic card represents x²+bx+c with negative discriminant.
• Repeated factors are entered through multiplicity, not by duplicating factor cards.
• The factor and degree limits keep the system readable for users.

Which templates are used?

The partial fraction template depends on the denominator factor type. Linear factors receive constant numerators; irreducible quadratic factors receive linear numerators.

• Distinct linear factor: A/(x-r).
• Repeated linear factor: A1/(x-r) + A2/(x-r)² + ... with one term for each power.
• Irreducible quadratic factor: (Ax+B)/(x²+bx+c).
• Repeated irreducible quadratic factor: one linear-numerator term for each supported power.

How are the coefficients found?

The coefficients are found by combining the partial fraction template over the common denominator and matching polynomial coefficients degree by degree. This produces a square linear system for the unknowns.

1. Build the full denominator D(x) from all entered factors.
2. For each unknown coefficient, compute its contribution using D(x) divided by the relevant factor power.
3. For quadratic terms, include both the x-part and the constant part of the numerator.
4. Match the resulting polynomial coefficients against the original numerator.
5. Solve the resulting linear system and substitute the coefficient values into the template.

The calculator is not just fitting sample points. It solves the coefficient identity, then uses numeric verification as an additional check for the displayed result.

Example: distinct linear factors

For distinct linear factors, each factor receives one constant numerator. For example, (5x+3)/((x-1)(x+2)) is decomposed as A/(x-1) + B/(x+2).

1. Clear denominators: A(x+2) + B(x-1) = 5x + 3.
2. Match coefficients: A+B = 5 and 2A-B = 3.
3. Solve the system: A = 8/3 and B = 7/3.
4. Result: (8/3)/(x-1) + (7/3)/(x+2).

Example: repeated linear factor

A repeated linear factor needs one term for each power. For (3x+5)/(x-1)², the correct template is A/(x-1) + B/(x-1)².

1. Clear denominators: A(x-1) + B = 3x + 5.
2. From the x coefficient, A = 3.
3. From the constant term, -A + B = 5, so B = 8.
4. Result: 3/(x-1) + 8/(x-1)².

Example: irreducible quadratic factor

An irreducible quadratic denominator uses a linear numerator. For example, (2x²+3x+5)/((x-1)(x²+1)) is decomposed as A/(x-1) + (Bx+C)/(x²+1).

1. Clear denominators: A(x²+1) + (Bx+C)(x-1) = 2x² + 3x + 5.
2. Match coefficients: A+B = 2, C-B = 3, and A-C = 5.
3. Solve the system: A = 5, B = -3, C = 0.
4. Result: 5/(x-1) - 3x/(x²+1).

How to read the verification badge

The verification badge means the decomposed expression matched the original rational function at several sample x-values that do not make the denominator zero.

This is a useful cross-check for the displayed result. The main solution still comes from coefficient matching and solving the linear system; the badge is not a replacement for the symbolic identity.

Common mistakes

The most common mistakes are entering the denominator structure incorrectly or trying to decompose an improper rational function directly.

• Trying partial fractions before polynomial division when the numerator degree is at least the denominator degree.
• Entering a reducible quadratic such as x²-5x+6 as an irreducible quadratic.
• Duplicating a factor card instead of using multiplicity.
• Forgetting that repeated factors require terms for each power.
• Treating numeric verification as the whole proof rather than a cross-check.

Limitations of this calculator

This calculator works with structured denominator factors; it does not auto-factor arbitrary denominator polynomials. That limitation is intentional because safe general factorization would require a broader symbolic algebra system.

• The total denominator degree is limited to 6.
• The factor list is limited to 4 cards.
• Linear factor multiplicity is supported up to 3.
• Irreducible quadratic multiplicity is supported up to 2.
• Improper fractions are rejected until polynomial division is performed first.
• The calculator is not a full symbolic algebra system or automatic polynomial factorization tool.