What Is Partial Fraction Decomposition?
Partial fraction decomposition rewrites a rational function as a sum of simpler fractions whose denominators are the factors of the original denominator. This makes it possible to integrate rational functions in calculus, and simplifies algebraic manipulation in precalculus.
P(x) / Q(x) = A/(factor 1) + B/(factor 2) + ...
where Q(x) is factored completely over the reals
When to Use It — Decision Checklist
Is the fraction proper? (degree of numerator < degree of denominator)
Is the denominator fully factored?
Are all factors distinct linear factors?
Step 1: Ensure a Proper Fraction
Partial fraction decomposition only applies to proper fractions — fractions where the degree of the numerator is strictly less than the degree of the denominator. If your fraction is improper, divide first.
Proper — Ready to Decompose
(3x + 1) / (x² − 4)
deg(num) = 1 < deg(den) = 2 ✓
Improper — Divide First
(x³ + 2x) / (x² − 1)
deg(num) = 3 ≥ deg(den) = 2 — do long division first
Long Division Example
Divide (x³ + 2x) ÷ (x² − 1) before decomposing.
x³ ÷ x² = x → x(x² − 1) = x³ − x
Subtract: (x³ + 2x) − (x³ − x) = 3x
3x ÷ x² → remainder (degree 1 < degree 2)
Now decompose only the proper fraction part: 3x/(x² − 1).
The Four Cases: Setup Formulas
After factoring the denominator completely, identify each factor type and write the corresponding partial fraction term(s).
Distinct Linear Factors
Each linear factor appears exactly once
Example denominator
Denominator: (x − 1)(x + 3)(2x − 5)
Partial fraction setup
A/(x − 1) + B/(x + 3) + C/(2x − 5)
Key rule: One constant per factor. Cover-up method works perfectly here.
Repeated Linear Factors
A linear factor appears with multiplicity > 1
Example denominator
Denominator: (x − 2)² or (x + 1)³
Partial fraction setup
A/(x − 2) + B/(x − 2)² [for (x−2)²]
Key rule: One term for each power from 1 up to the multiplicity.
Irreducible Quadratic Factors
Quadratic factor with discriminant b²−4ac < 0
Example denominator
Denominator: (x² + 4) or (x² + x + 1)
Partial fraction setup
(Ax + B)/(x² + 4)
Key rule: Numerator must be linear (Ax+B), not just a constant.
Repeated Irreducible Quadratic Factors
Irreducible quadratic appears with multiplicity > 1
Example denominator
Denominator: (x² + 1)²
Partial fraction setup
(Ax + B)/(x² + 1) + (Cx + D)/(x² + 1)²
Key rule: One linear-numerator term for each power of the quadratic.
The Cover-Up Method (Heaviside's Method)
When all denominator factors are distinct linear factors, the cover-up method finds each constant in one step — no algebra needed. For a fraction P(x)/[(x−a)(x−b)(x−c)...], to find the constant over (x−a): cover up (x−a) in the denominator and evaluate what remains at x = a.
Cover-Up Rule
To find A in A/(x−a): evaluate P(x)/[all other factors] at x = a.
Cover-Up Walkthrough
Decompose: 5x / [(x − 1)(x + 2)(x − 3)]
Set up: 5x / [(x−1)(x+2)(x−3)] = A/(x−1) + B/(x+2) + C/(x−3)
Find A: Cover (x−1). Evaluate 5x/[(x+2)(x−3)] at x = 1.
5(1) / [(1+2)(1−3)] = 5 / [3·(−2)] = 5/(−6) = −5/6
A = −5/6
Find B: Cover (x+2). Evaluate 5x/[(x−1)(x−3)] at x = −2.
5(−2) / [(−2−1)(−2−3)] = −10 / [(−3)(−5)] = −10/15 = −2/3
B = −2/3
Find C: Cover (x−3). Evaluate 5x/[(x−1)(x+2)] at x = 3.
5(3) / [(3−1)(3+2)] = 15 / [2·5] = 15/10 = 3/2
C = 3/2
Result: −5/6 · 1/(x−1) − 2/3 · 1/(x+2) + 3/2 · 1/(x−3)
Equating Coefficients Method
For repeated or quadratic factors, set up the partial fraction form, multiply both sides by the denominator to clear fractions, then either substitute strategic x-values or expand and match coefficients of each power of x.
Write the partial fraction setup with unknown constants A, B, C, ...
Multiply both sides by the full denominator Q(x) to clear all fractions.
Substitute roots of each linear factor to find constants (substitution method).
For remaining unknowns, expand the right side and match coefficients of like powers of x.
Solve the resulting system of equations for all constants.
4 Fully Worked Examples
Example 1 — Distinct Linear Factors
Decompose: (2x + 3) / [(x − 1)(x + 2)]
Step 1 — Check degrees & write setup
deg(num) = 1 < deg(den) = 2 ✓ (proper)
(2x + 3)/[(x−1)(x+2)] = A/(x−1) + B/(x+2)
Step 2 — Multiply through by (x−1)(x+2)
2x + 3 = A(x + 2) + B(x − 1)
Step 3 — Substitute roots (cover-up)
x = 1: 2(1) + 3 = A(1+2) + B(0) → 5 = 3A → A = 5/3
x = −2: 2(−2) + 3 = A(0) + B(−2−1) → −1 = −3B → B = 1/3
(2x + 3) / [(x−1)(x+2)] = (5/3)/(x−1) + (1/3)/(x+2)
Verify: combine the two fractions and confirm you get back (2x+3)/[(x−1)(x+2)].
Example 2 — Repeated Linear Factor
Decompose: (3x − 1) / [x(x − 2)²]
Step 1 — Check & write setup
deg(num) = 1 < deg(den) = 3 ✓
(3x−1) / [x(x−2)²] = A/x + B/(x−2) + C/(x−2)²
Note: (x−2)² requires two terms — one for each power.
Step 2 — Multiply through by x(x−2)²
3x − 1 = A(x−2)² + Bx(x−2) + Cx
Step 3 — Substitution
x = 0: −1 = A(4) + 0 + 0 → A = −1/4
x = 2: 6−1 = 0 + 0 + C(2) → 5 = 2C → C = 5/2
Step 4 — Equate coefficients to find B
Expand right: A(x²−4x+4) + B(x²−2x) + Cx
x² coefficient: 0 = A + B → B = −A = 1/4 → B = 1/4
(3x−1) / [x(x−2)²] = −(1/4)/x + (1/4)/(x−2) + (5/2)/(x−2)²
Example 3 — Irreducible Quadratic Factor
Decompose: (x² + 5x + 2) / [(x − 1)(x² + 4)]
Step 1 — Identify factor types & write setup
x² + 4 has discriminant 0 − 16 = −16 < 0 → irreducible quadratic
(x²+5x+2)/[(x−1)(x²+4)] = A/(x−1) + (Bx+C)/(x²+4)
The irreducible quadratic requires a linear numerator Bx+C.
Step 2 — Multiply through by (x−1)(x²+4)
x² + 5x + 2 = A(x²+4) + (Bx+C)(x−1)
Step 3 — Substitute x = 1 for A
1 + 5 + 2 = A(1+4) + 0 → 8 = 5A → A = 8/5
Step 4 — Equate coefficients for B and C
Expand right: Ax²+4A + Bx²−Bx+Cx−C
= (A+B)x² + (C−B)x + (4A−C)
x² coeff: 1 = A + B → B = 1 − 8/5 = −3/5
constant: 2 = 4A − C → C = 4(8/5) − 2 = 32/5 − 10/5 = 22/5
= (8/5)/(x−1) + (−3x/5 + 22/5)/(x²+4)
= (8/5)/(x−1) + (−3x + 22) / [5(x²+4)]
Example 4 — Improper Fraction (Long Division First)
Decompose: (x³ + x² − 2) / [(x − 1)(x + 1)]
Step 1 — Check degrees
deg(num) = 3 > deg(den) = 2 → improper. Must do long division first.
Denominator = (x−1)(x+1) = x² − 1
Step 2 — Long division: (x³ + x² − 2) ÷ (x² − 1)
x³ ÷ x² = x → x(x²−1) = x³ − x
Subtract: (x³+x²−2) − (x³−x) = x²+x−2
x² ÷ x² = 1 → 1(x²−1) = x²−1
Subtract: (x²+x−2) − (x²−1) = x − 1
(x³+x²−2)/(x²−1) = (x + 1) + (x−1)/(x²−1)
Step 3 — Simplify the remainder fraction
(x−1)/[(x−1)(x+1)] = 1/(x+1) [cancel (x−1), x ≠ 1]
(x³+x²−2) / [(x−1)(x+1)] = x + 1 + 1/(x+1)
The remainder simplified to 1/(x+1) — no further partial fraction work needed.
Why This Matters: Connection to Integration
In calculus, partial fractions transform difficult integrals into sums of standard forms. Each type of partial fraction has a known integral formula.
| Partial Fraction Form | Integral |
|---|---|
| A/(x − a) | A · ln|x − a| + C |
| A/(x − a)² | −A/(x − a) + C |
| A/(x − a)ⁿ (n ≥ 2) | A · (x−a)^(1−n) / (1−n) + C |
| (Ax + B)/(x² + k²) | A/2 · ln(x²+k²) + B/k · arctan(x/k) + C |
Integration Example
∫ (2x + 3) / [(x−1)(x+2)] dx
From Example 1: = ∫ [(5/3)/(x−1) + (1/3)/(x+2)] dx
= 5/3 · ln|x−1| + 1/3 · ln|x+2| + C
Common Mistakes to Avoid
Not checking if the fraction is proper
Fix: Always verify deg(numerator) < deg(denominator). If not, do polynomial long division first.
Missing terms for repeated factors
Fix: (x−a)² needs both A/(x−a) AND B/(x−a)². Never skip the intermediate power.
Using a constant numerator for a quadratic factor
Fix: An irreducible quadratic factor x²+bx+c requires a linear numerator (Ax+B), not just A.
Sign errors when multiplying through
Fix: Distribute carefully. Write out each expansion fully before collecting terms.
Incomplete factoring of the denominator
Fix: Factor Q(x) completely. A quadratic that can be factored over the reals must be factored.
Applying cover-up to repeated or quadratic factors
Fix: Cover-up only works for simple (non-repeated) linear factors. Use equating coefficients otherwise.
Practice Problems
Try these on your own before checking the answers. Each covers a different case.
Decompose: 7 / [(x − 3)(x + 4)]
Hint: Case 1 — distinct linear factors. Use cover-up.
Decompose: (4x + 1) / [x²(x − 1)]
Hint: Case 2 — repeated linear factor x² needs A/x + B/x².
Decompose: (2x² − x + 4) / [(x + 1)(x² + 3)]
Hint: Case 3 — irreducible quadratic x²+3. Setup: A/(x+1) + (Bx+C)/(x²+3).
Decompose: (x³ + 4x) / (x² + 4)
Hint: Check degrees first — improper fraction. Do long division, then decompose.
Quick Reference Card
Frequently Asked Questions
What is partial fraction decomposition?
Partial fraction decomposition is the process of writing a rational expression (a fraction of polynomials) as a sum of simpler fractions. For example, 5/(x²−1) can be written as 5/[(x−1)(x+1)] = A/(x−1) + B/(x+1). This technique is essential for integrating rational functions in calculus and appears in precalculus when working with rational expressions.
When do you need to do polynomial long division first?
You must perform polynomial long division first whenever the degree of the numerator is greater than or equal to the degree of the denominator — this is called an improper fraction. For example, (x³ + 2x)/(x² − 1) is improper because the numerator degree (3) is not less than the denominator degree (2). After long division, you get a polynomial plus a proper fraction remainder. Only the proper fraction part undergoes partial fraction decomposition.
What is the cover-up method for partial fractions?
The cover-up method (also called Heaviside's method) is a shortcut for finding constants in partial fractions when you have distinct linear factors. To find A in A/(x−a), mentally cover up the factor (x−a) in the original denominator, then evaluate the rest of the fraction at x = a. For example, to decompose 5/[(x−1)(x+1)]: cover (x−1) and evaluate 5/(x+1) at x=1 to get A = 5/2. Cover (x+1) and evaluate 5/(x−1) at x=−1 to get B = −5/2.
How do you handle repeated linear factors in partial fractions?
When the denominator has a repeated linear factor (ax+b)^n, you must include one partial fraction term for each power from 1 to n. For example, if the denominator contains (x−2)², you need two terms: A/(x−2) + B/(x−2)². For (x−2)³, you need A/(x−2) + B/(x−2)² + C/(x−2)³. You cannot determine all constants by the cover-up method alone — use a combination of substitution and equating coefficients.
What is an irreducible quadratic factor?
An irreducible quadratic factor is a quadratic ax²+bx+c that cannot be factored over the real numbers, meaning its discriminant b²−4ac is negative. For example, x²+4 and x²+x+1 are irreducible. When such a factor appears in the denominator, the corresponding partial fraction numerator must be a linear expression (Ax+B), not just a constant. So for a term like (x²+4), you write (Ax+B)/(x²+4).
Why do partial fractions matter in calculus?
Partial fraction decomposition is the gateway to integrating rational functions. Integrals like ∫ 1/(x²−1) dx are difficult to compute directly, but after decomposing to ∫ [1/2·1/(x−1) − 1/2·1/(x+1)] dx, each term integrates easily using ∫ 1/(x−a) dx = ln|x−a| + C. The technique also appears in differential equations (Laplace transforms) and in solving certain series problems.
How do you set up partial fractions for a mix of factor types?
Write one partial fraction term for each factor in the denominator, following the rules for each type: (1) Distinct linear factor (ax+b): constant numerator A/(ax+b). (2) Repeated linear factor (ax+b)^n: one term per power, A/(ax+b) through A_n/(ax+b)^n. (3) Irreducible quadratic (ax²+bx+c): linear numerator (Ax+B)/(ax²+bx+c). (4) Repeated irreducible quadratic (ax²+bx+c)^n: one term per power with linear numerators. The total number of unknown constants must equal the degree of the denominator.
What are the most common mistakes in partial fraction decomposition?
The five most common mistakes are: (1) Forgetting to check if the fraction is improper — always verify degree(numerator) < degree(denominator) before starting. (2) Missing a repeated factor term — (x−2)² requires both A/(x−2) and B/(x−2)². (3) Using a constant numerator for an irreducible quadratic — it must be (Ax+B)/(x²+bx+c). (4) Sign errors when multiplying through by the denominator — be especially careful with minus signs. (5) Failing to fully factor the denominator — an unfactored denominator leads to the wrong decomposition form.
Related Topics
Polynomial Long Division
Master long division and synthetic division — prerequisite for partial fractions.
Rational Functions
Asymptotes, holes, domain, and graphing rational expressions.
Polynomials
Factoring, zeros, and polynomial behavior — essential for decomposing denominators.
Precalculus Study Guide
Complete coverage of all precalculus topics.
Sequences & Series
Partial fractions appear in telescoping series problems.
Practice Problems
Step-by-step solutions with AI explanations.
Practice Partial Fractions
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