Partial Fraction Decomposition

Partial fraction decomposition is the reverse of adding fractions. Just as you combine $\frac{1}{x-1} + \frac{2}{x+3}$ into a single fraction, partial fractions breaks $\frac{5x + 7}{(x-1)(x+3)}$ back into a sum of simpler fractions. This technique is essential for integration in calculus, for inverse Laplace transforms in engineering, and for simplifying complex rational expressions.

Why Partial Fractions?

Consider the fraction $\frac{5x + 7}{x^2 + 2x - 3}$. As a single fraction, it is hard to integrate, hard to analyze, and hard to invert in a Laplace transform. But if we write it as:

$\frac{5x + 7}{x^2 + 2x - 3} = \frac{3}{x - 1} + \frac{2}{x + 3}$

Each piece has a simple, well-known integral ($\ln|x-1|$ and $\ln|x+3|$), and each piece has a straightforward Laplace inverse. The decomposition makes hard problems easy.

The Setup: Three Cases

The method depends on the type of factors in the denominator after complete factoring.

Case 1: Distinct Linear Factors

If the denominator factors into distinct (non-repeating) linear factors:

$\frac{P(x)}{(x - a_1)(x - a_2) \cdots (x - a_n)} = \frac{A_1}{x - a_1} + \frac{A_2}{x - a_2} + \cdots + \frac{A_n}{x - a_n}$

Each factor gets one constant in the numerator.

Example 1: Decompose $\frac{5x + 7}{(x - 1)(x + 3)}$.

Setup:

$\frac{5x + 7}{(x - 1)(x + 3)} = \frac{A}{x - 1} + \frac{B}{x + 3}$

Solve: Multiply both sides by $(x - 1)(x + 3)$:

$5x + 7 = A(x + 3) + B(x - 1)$

Strategic substitution — choose $x$-values that eliminate one variable:

Set $x = 1$: $5(1) + 7 = A(1 + 3) + B(0) \implies 12 = 4A \implies A = 3$

Set $x = -3$: $5(-3) + 7 = A(0) + B(-3 - 1) \implies -8 = -4B \implies B = 2$

Result:

$\frac{5x + 7}{(x - 1)(x + 3)} = \frac{3}{x - 1} + \frac{2}{x + 3}$

Verification: $\frac{3}{x-1} + \frac{2}{x+3} = \frac{3(x+3) + 2(x-1)}{(x-1)(x+3)} = \frac{3x+9+2x-2}{(x-1)(x+3)} = \frac{5x+7}{(x-1)(x+3)}$ — confirmed.

Case 2: Repeated Linear Factors

If a linear factor $(x - a)$ appears with multiplicity $k$, you need $k$ terms with increasing powers:

$\frac{P(x)}{(x - a)^k} = \frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + \cdots + \frac{A_k}{(x - a)^k}$

Example 2: Decompose $\frac{3x^2 + 5x + 2}{(x + 1)^2(x - 2)}$.

Setup:

$\frac{3x^2 + 5x + 2}{(x + 1)^2(x - 2)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x - 2}$

Solve: Multiply both sides by $(x + 1)^2(x - 2)$:

$3x^2 + 5x + 2 = A(x + 1)(x - 2) + B(x - 2) + C(x + 1)^2$

Set $x = 2$: $3(4) + 5(2) + 2 = C(3)^2 \implies 24 = 9C \implies C = \frac{8}{3}$

Set $x = -1$: $3(1) + 5(-1) + 2 = B(-3) \implies 0 = -3B \implies B = 0$

Set $x = 0$ (using known $B$ and $C$): $2 = A(1)(-2) + 0(-2) + \frac{8}{3}(1) \implies 2 = -2A + \frac{8}{3} \implies -2A = 2 - \frac{8}{3} = -\frac{2}{3} \implies A = \frac{1}{3}$

Result:

$\frac{3x^2 + 5x + 2}{(x + 1)^2(x - 2)} = \frac{1/3}{x + 1} + \frac{0}{(x + 1)^2} + \frac{8/3}{x - 2} = \frac{1}{3(x + 1)} + \frac{8}{3(x - 2)}$

Case 3: Irreducible Quadratic Factors

An irreducible quadratic is a quadratic $ax^2 + bx + c$ that cannot be factored over the real numbers (its discriminant $b^2 - 4ac < 0$). For each irreducible quadratic factor, the partial fraction has a linear numerator:

$\frac{P(x)}{(ax^2 + bx + c)} \to \frac{Ax + B}{ax^2 + bx + c}$

If the irreducible quadratic is repeated with multiplicity $k$, you need $k$ terms:

$\frac{A_1 x + B_1}{ax^2 + bx + c} + \frac{A_2 x + B_2}{(ax^2 + bx + c)^2} + \cdots$

Example 3: Decompose $\frac{2x^2 + x + 3}{(x - 1)(x^2 + 1)}$.

Note: $x^2 + 1$ is irreducible (discriminant $0 - 4 = -4 < 0$).

Setup:

$\frac{2x^2 + x + 3}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$

Solve: Multiply both sides by $(x - 1)(x^2 + 1)$:

$2x^2 + x + 3 = A(x^2 + 1) + (Bx + C)(x - 1)$

Set $x = 1$: $2 + 1 + 3 = A(2) \implies 6 = 2A \implies A = 3$

Expand the right side: $A(x^2 + 1) + (Bx + C)(x - 1) = Ax^2 + A + Bx^2 - Bx + Cx - C$

$= (A + B)x^2 + (-B + C)x + (A - C)$

Match coefficients with $2x^2 + x + 3$:

• $x^2$: $A + B = 2 \implies 3 + B = 2 \implies B = -1$
• $x^1$: $-B + C = 1 \implies 1 + C = 1 \implies C = 0$
• $x^0$: $A - C = 3 \implies 3 - 0 = 3$ — confirmed

Result:

$\frac{2x^2 + x + 3}{(x - 1)(x^2 + 1)} = \frac{3}{x - 1} + \frac{-x}{x^2 + 1}$

The General Procedure

1. Ensure the fraction is proper (numerator degree less than denominator degree). If not, perform long division first.
2. Factor the denominator completely.
3. Set up the decomposition according to the three cases above.
4. Clear denominators by multiplying both sides by the full denominator.
5. Solve for coefficients using strategic substitution (plug in roots of denominator factors) and/or coefficient matching.
6. Verify by recombining the partial fractions.

Important: The Fraction Must Be Proper

If $\deg(P) \geq \deg(Q)$, you must divide first.

Example 4: Decompose $\frac{x^3 + 2x^2 + x + 1}{x^2 - 1}$.

Numerator degree 3 is greater than denominator degree 2. Divide first:

$x^3 + 2x^2 + x + 1 \div (x^2 - 1)$

Step 1: $x^3 \div x^2 = x$. Multiply: $x^3 - x$. Subtract: $2x^2 + 2x + 1$.

Step 2: $2x^2 \div x^2 = 2$. Multiply: $2x^2 - 2$. Subtract: $2x + 3$.

$\frac{x^3 + 2x^2 + x + 1}{x^2 - 1} = x + 2 + \frac{2x + 3}{x^2 - 1}$

Now decompose the proper fraction $\frac{2x + 3}{(x-1)(x+1)}$:

$\frac{2x + 3}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$

$x = 1$: $5 = 2A \implies A = \frac{5}{2}$

$x = -1$: $1 = -2B \implies B = -\frac{1}{2}$

Final result:

$\frac{x^3 + 2x^2 + x + 1}{x^2 - 1} = x + 2 + \frac{5/2}{x - 1} - \frac{1/2}{x + 1}$

Solving Systems by Coefficient Matching

When strategic substitution does not give you all constants (common with irreducible quadratics), expand the right side and match coefficients of like powers of $x$.

Example 5: Decompose $\frac{x + 4}{(x^2 + 4)(x - 1)}$.

Setup:

$\frac{x + 4}{(x^2 + 4)(x - 1)} = \frac{Ax + B}{x^2 + 4} + \frac{C}{x - 1}$

Multiply through: $x + 4 = (Ax + B)(x - 1) + C(x^2 + 4)$

Set $x = 1$: $5 = C(5) \implies C = 1$

Expand: $(Ax + B)(x-1) + (x^2 + 4) = Ax^2 - Ax + Bx - B + x^2 + 4$

$= (A + 1)x^2 + (-A + B)x + (-B + 4)$

Match with $0 \cdot x^2 + 1 \cdot x + 4$:

• $x^2$: $A + 1 = 0 \implies A = -1$
• $x^1$: $-A + B = 1 \implies 1 + B = 1 \implies B = 0$
• $x^0$: $-B + 4 = 4 \implies 4 = 4$ — confirmed

Result:

$\frac{x + 4}{(x^2 + 4)(x - 1)} = \frac{-x}{x^2 + 4} + \frac{1}{x - 1}$

Real-World Application: Electrical Circuit Analysis

In electrical engineering, the transfer function of a circuit is often a rational function of the complex frequency variable $s$. To find the time-domain response, engineers use inverse Laplace transforms, which require partial fraction decomposition.

For example, a circuit might produce:

$H(s) = \frac{10}{(s + 2)(s + 5)}$

Decomposing: $\frac{10}{(s+2)(s+5)} = \frac{A}{s+2} + \frac{B}{s+5}$

$s = -2$: $10 = 3A \implies A = \frac{10}{3}$

$s = -5$: $10 = -3B \implies B = -\frac{10}{3}$

$H(s) = \frac{10/3}{s + 2} - \frac{10/3}{s + 5}$

Each term has a known inverse: $\frac{a}{s + b} \leftrightarrow ae^{-bt}$. So the time response is:

$h(t) = \frac{10}{3}e^{-2t} - \frac{10}{3}e^{-5t}$

Without partial fractions, finding this inverse would be far more difficult.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Partial fraction decomposition breaks a complex rational expression into a sum of simpler fractions
• Distinct linear factors each get one constant: $\frac{A}{x - a}$
• Repeated linear factors need one term per power: $\frac{A}{x-a} + \frac{B}{(x-a)^2} + \cdots$
• Irreducible quadratic factors get a linear numerator: $\frac{Ax + B}{ax^2 + bx + c}$
• Always check that the fraction is proper first — if not, divide before decomposing
• Use strategic substitution (roots of denominator) for quick solutions, and coefficient matching when substitution is not enough
• The technique is foundational for calculus integration and Laplace transforms in engineering

Return to College Algebra for more topics in this section.