9.4Decomposing Rational Expressions
Decompose P(x)/Q(x) into partial fractions for distinct linear factors, repeated linear factors, and irreducible quadratic factors. Set up and solve the resulting systems of equations.
Partial fraction decomposition is the key technique for integrating rational functions in calculus. Every integral of the form ∫ P(x)/Q(x) dx relies on this decomposition.
Essential Question
How does partial fraction decomposition reverse the process of adding rational expressions, and why is this technique essential for integration in calculus?
Lesson Overview
Partial fraction decomposition is the process of writing a rational expression P(x)/Q(x) as a sum of simpler fractions. It is the reverse of adding rational expressions.
⚠️ Prerequisite: The degree of the numerator must be LESS than the degree of the denominator. If not, perform polynomial long division first.
Case 1 — Distinct Linear Factors
Q(x) = (x−a)(x−b)···
P(x)/Q(x) = A/(x−a) + B/(x−b) + ···
Case 2 — Repeated Linear Factors
Q(x) = (x−a)ⁿ
Include: A₁/(x−a) + A₂/(x−a)² + ··· + Aₙ/(x−a)ⁿ
Case 3 — Irreducible Quadratic Factors
Q(x) = (ax²+bx+c) where b²−4ac < 0
Include: (Ax+B)/(ax²+bx+c)
Case 4 — Repeated Irreducible Quadratic Factors
Include: (A₁x+B₁)/(ax²+bx+c) + (A₂x+B₂)/(ax²+bx+c)² + ···
Methods to Find Constants
Method 1 (Plug in roots): Multiply both sides by Q(x), then substitute each root of Q(x) to solve for the constants directly.
Method 2 (Coefficient matching): Expand both sides and equate coefficients of like powers of x to form a system of equations.
Worked Examples
Decompose: (3x+5) / ((x+1)(x+2))
Set up: (3x+5)/((x+1)(x+2)) = A/(x+1) + B/(x+2)
Multiply both sides by (x+1)(x+2):
3x + 5 = A(x+2) + B(x+1)
Plug in x = −1: 3(−1)+5 = A(1) → 2 = A → A = 2
Plug in x = −2: 3(−2)+5 = B(−1) → −1 = −B → B = 1
Decompose: (2x²+3x−1) / ((x−1)(x+1)(x+2))
Set up: = A/(x−1) + B/(x+1) + C/(x+2)
Multiply: 2x²+3x−1 = A(x+1)(x+2) + B(x−1)(x+2) + C(x−1)(x+1)
x = 1: 2+3−1 = A(2)(3) → 4 = 6A → A = 2/3
x = −1: 2−3−1 = B(−2)(1) → −2 = −2B → B = 1
x = −2: 8−6−1 = C(−3)(−1) → 1 = 3C → C = 1/3
Decompose: (x+3) / (x(x−1)²)
Set up: = A/x + B/(x−1) + C/(x−1)² [repeated linear factor]
Multiply: x+3 = A(x−1)² + Bx(x−1) + Cx
x = 0: 3 = A(1) → A = 3
x = 1: 4 = C(1) → C = 4
Expand and match x² coefficient: 0 = A + B → B = −3
Decompose: (2x+1) / ((x²+1)(x−1))
Set up: = (Ax+B)/(x²+1) + C/(x−1) [irreducible quadratic + linear]
Multiply: 2x+1 = (Ax+B)(x−1) + C(x²+1)
x = 1: 3 = 2C → C = 3/2
Expand: (A+C)x² + (−A+B)x + (−B+C) = 2x+1
Match x²: A + C = 0 → A = −3/2
Match x¹: −A + B = 2 → B = 2 − 3/2 = 1/2
Check x⁰: −B + C = −1/2 + 3/2 = 1 ✓
Decompose: (x²+2x+3) / ((x+1)(x²+2x+2))
Note: x²+2x+2 is irreducible (discriminant = 4−8 = −4 < 0)
Set up: = A/(x+1) + (Bx+C)/(x²+2x+2)
Multiply: x²+2x+3 = A(x²+2x+2) + (Bx+C)(x+1)
x = −1: 1−2+3 = A(1−2+2) → 2 = A
Expand: (A+B)x² + (2A+B+C)x + (2A+C) = x²+2x+3
Match x²: A+B = 1 → B = −1
Match x¹: 2A+B+C = 2 → 4−1+C = 2 → C = −1
Check x⁰: 2A+C = 4−1 = 3 ✓
Guided Practice
Decompose: (5x+1) / ((x+1)(x−1))
Hint: Set up A/(x+1) + B/(x−1), multiply through, and plug in x = −1 and x = 1.
Decompose: (3x+7) / ((x+2)²)
Hint: Repeated linear factor — use A/(x+2) + B/(x+2)².
Decompose: (x²+1) / (x(x+1)(x−2))
Hint: Three distinct linear factors — use A/x + B/(x+1) + C/(x−2).
Decompose: (2x+3) / ((x²+1)(x+1))
Hint: One irreducible quadratic and one linear factor — use (Ax+B)/(x²+1) + C/(x+1).
Decompose: (x³+2x²+x+1) / (x(x²+1))
Hint: Degree of numerator = degree of denominator — do polynomial long division first, then decompose the remainder.
Key Vocabulary
Partial fraction decomposition
Writing P(x)/Q(x) as a sum of simpler rational expressions with lower-degree denominators.
Example: (3x+5)/((x+1)(x+2)) = 2/(x+1) + 1/(x+2)
Proper rational expression
A rational expression where the degree of the numerator is strictly less than the degree of the denominator.
Example: (2x+1)/(x²+3x+2) — numerator degree 1 < denominator degree 2
Improper rational expression
A rational expression where the degree of the numerator is greater than or equal to the degree of the denominator. Requires polynomial long division first.
Example: (x³+1)/(x²+x) — numerator degree 3 ≥ denominator degree 2
Distinct linear factors
Factors of the form (x−a) where all roots are different. Each contributes one constant A/(x−a) to the decomposition.
Repeated linear factors
A factor (x−a)ⁿ with multiplicity n > 1. Requires terms A₁/(x−a) + A₂/(x−a)² + ··· + Aₙ/(x−a)ⁿ.
Example: 1/(x−2)³ → A/(x−2) + B/(x−2)² + C/(x−2)³
Irreducible quadratic factor
A quadratic ax²+bx+c with discriminant b²−4ac < 0 — it cannot be factored over the real numbers. Requires a linear numerator Ax+B.
Example: x²+4 is irreducible since discriminant = 0−16 < 0
Coefficient matching
Expanding both sides of an equation and equating coefficients of like powers of x to form a system of equations for the unknown constants.
Check Your Understanding
Interactive Practice — 5 Questions
The partial fraction form of A/((x+1)(x−2)) uses:
Before decomposing (x³+1)/(x²+x), you must first:
The partial fraction form for 1/((x−1)²(x+2)) is:
For an irreducible quadratic factor (x²+4), the partial fraction term is:
Decompose 1/((x+1)(x−1)). The constants A and B are:
Common Mistakes
Using a constant numerator A for an irreducible quadratic factor
Irreducible quadratic factors (x²+bx+c) require a LINEAR numerator Ax+B, not just A.
Forgetting to include all powers for repeated factors
For (x−a)², you need BOTH A/(x−a) AND B/(x−a)². Include every power from 1 up to n.
Trying to decompose an improper fraction directly
If degree(numerator) ≥ degree(denominator), do polynomial long division FIRST, then decompose the remainder.
Only using the 'plug in roots' method and missing constants for irreducible quadratics
For irreducible quadratic factors, plug in roots won't work (they're complex). Use coefficient matching instead.
Math Tips
The 'plug in roots' method is the fastest way to find constants for linear factors. For each factor (x−a), plug in x=a.
After finding constants by plugging in roots, use coefficient matching to find any remaining constants (for irreducible quadratic factors).
Always check your answer by recombining the partial fractions — you should get back the original expression.
The number of unknown constants equals the degree of the denominator (after factoring). This is a useful check.
Partial fractions are the key technique for integrating rational functions in calculus. Mastering this now pays dividends later.