This page covers partial fraction decomposition from the ground up. The first section states the rules: how to check degrees, how to factor the denominator, and which numerator to use for each factor type. The second section then provides 7 fully solved practice problems covering distinct linear factors, repeated linear factors, and irreducible quadratic factors. Ideal for algebra, precalculus, and calculus students.
Partial fraction decomposition is the reverse of adding rational expressions. Given a rational function \[ \frac{P(x)}{Q(x)} \] with \( \deg P < \deg Q \), we rewrite it as a sum of simpler fractions whose denominators are the factors of \( Q(x) \).
If \( \deg P \ge \deg Q \), first perform polynomial long division and decompose the remainder term. The rules below only apply when \( \deg P < \deg Q \).
Factor \( Q(x) \) into linear factors and irreducible quadratic factors over the reals. Each distinct factor type dictates a different setup.
| Factor in \(Q(x)\) | Contribution to the decomposition |
|---|---|
| Distinct linear: \(x - a\) | \(\dfrac{A}{x - a}\) |
| Repeated linear: \((x - a)^n\) | \(\dfrac{A_1}{x-a} + \dfrac{A_2}{(x-a)^2} + \cdots + \dfrac{A_n}{(x-a)^n}\) |
| Irreducible quadratic: \(x^2 + bx + c\) | \(\dfrac{Bx + C}{x^2 + bx + c}\) |
| Repeated irreducible quadratic: \((x^2 + bx + c)^n\) | \(\dfrac{B_1 x + C_1}{x^2+bx+c} + \cdots + \dfrac{B_n x + C_n}{(x^2+bx+c)^n}\) |
In every case, the numerator has degree one less than the corresponding factor. A linear factor → constant numerator. A quadratic factor → linear numerator.
Two standard strategies:
Combine the resulting fractions and confirm the numerator matches \( P(x) \). Checking at one or two convenient values of \(x\) (e.g., \(x = 0\)) is a quick sanity test.
For \(\dfrac{5x + 10}{x(x+5)}\), the denominator has distinct linear factors \(x\) and \(x+5\), so we write: \[ \frac{5x + 10}{x(x+5)} = \frac{A}{x} + \frac{B}{x + 5} \] Substituting \(x = 0\) gives \(A = 2\); substituting \(x = -5\) gives \(B = 3\). This is the same problem as Question 1 below.
Decompose into partial fractions: \( \dfrac{5x + 10}{x(x + 5)} \)
Set up the decomposition:
\[ \frac{5x+10}{x(x+5)} = \frac{A}{x} + \frac{B}{x+5} \]Multiply both sides by \(x(x+5)\):
\[ 5x + 10 = A(x+5) + Bx \]At \(x=0\): \(10 = 5A \Rightarrow A = 2\).
At \(x=-5\): \(-15 = -5B \Rightarrow B = 3\).
Answer: \( \dfrac{2}{x} + \dfrac{3}{x+5} \)
Decompose into partial fractions: \( \dfrac{8x + 14}{(x + 1)(x + 5)} \)
At \(x=-1\): \(6 = 4A \Rightarrow A = \tfrac{3}{2}\).
At \(x=-5\): \(-26 = -4B \Rightarrow B = \tfrac{13}{2}\).
Answer: \( \dfrac{3}{2(x+1)} + \dfrac{13}{2(x+5)} \), equivalently \( \dfrac{1}{2}\!\left(\dfrac{3}{x+1} + \dfrac{13}{x+5}\right) \)
Decompose into partial fractions: \( \dfrac{5x^2 + 12x + 3}{x(x + 1)^2} \)
At \(x=0\): \(3 = A\).
At \(x=-1\): \(-4 = -C \Rightarrow C = 4\).
At \(x=1\): \(20 = 3(4) + 2B + 4 \Rightarrow B = 2\).
Answer: \( \dfrac{3}{x} + \dfrac{2}{x+1} + \dfrac{4}{(x+1)^2} \)
Decompose into partial fractions: \( \dfrac{3x + 15}{(x + 4)^2} \)
A quick trick avoids a full system of equations. Rewrite the numerator to contain \((x+4)\):
\[ 3x + 15 = 3(x+4) + 3 \]Split the fraction:
\[ \frac{3(x+4) + 3}{(x+4)^2} = \frac{3(x+4)}{(x+4)^2} + \frac{3}{(x+4)^2} = \frac{3}{x+4} + \frac{3}{(x+4)^2} \]Answer: \( \dfrac{3}{x+4} + \dfrac{3}{(x+4)^2} \)
Decompose into partial fractions: \( \dfrac{7x + 10}{(x + 1)(x^2 - 4)} \)
Factor \(x^2 - 4\) as a difference of squares:
\[ \frac{7x+10}{(x+1)(x-2)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2} \] \[ 7x + 10 = A(x-2)(x+2) + B(x+1)(x+2) + C(x+1)(x-2) \]At \(x=-1\): \(3 = -3A \Rightarrow A = -1\).
At \(x=2\): \(24 = 12B \Rightarrow B = 2\).
At \(x=-2\): \(-4 = 4C \Rightarrow C = -1\).
Answer: \( -\dfrac{1}{x+1} + \dfrac{2}{x-2} - \dfrac{1}{x+2} \)
Decompose into partial fractions: \( \dfrac{5x^2 + 31x + 46}{(x + 2)(x + 3)^2} \)
At \(x=-2\): \(4 = A\).
At \(x=-3\): \(-2 = -C \Rightarrow C = 2\).
Match \(x^2\) coefficients: \(A + B = 5 \Rightarrow B = 1\).
Answer: \( \dfrac{4}{x+2} + \dfrac{1}{x+3} + \dfrac{2}{(x+3)^2} \)
Decompose into partial fractions: \( \dfrac{2x^2 + 6x - 2}{x^3 - 1} \)
Factor the difference of cubes:
\[ x^3 - 1 = (x-1)(x^2 + x + 1) \]Since \(x^2+x+1\) is irreducible over \(\mathbb{R}\), set up:
\[ \frac{2x^2+6x-2}{(x-1)(x^2+x+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \] \[ 2x^2 + 6x - 2 = A(x^2 + x + 1) + (Bx + C)(x-1) \]At \(x=1\): \(6 = 3A \Rightarrow A = 2\).
After expanding and matching coefficients:
\[ 2x^2 + 6x - 2 = (2+B)x^2 + (2 - B + C)x + (2 - C) \]\(x^2\): \(2 = 2 + B \Rightarrow B = 0\).
Constant: \(-2 = 2 - C \Rightarrow C = 4\).
Answer: \( \dfrac{2}{x-1} + \dfrac{4}{x^2 + x + 1} \)
It's the reverse of adding rational expressions: given a rational function, break it into a sum of simpler fractions whose denominators are the factors of the original denominator.
Whenever the corresponding denominator factor is an irreducible quadratic (e.g., \(x^2+x+1\) or \(x^2+4\)). Linear denominators use a constant numerator; quadratics need a linear numerator.
Include one term for each power: \(\dfrac{A_1}{x-a} + \dfrac{A_2}{(x-a)^2} + \cdots + \dfrac{A_n}{(x-a)^n}\).
Perform polynomial long division first, then apply partial fraction decomposition to the remainder term.
It converts complicated rational functions into sums of simpler fractions, making integration, inverse Laplace transforms, and series expansions much easier.