Subject: Pre-Calculus, Calculus 2, and Differential Equations
Created by: Leo Perez
Revised:2/24/2018
Partial Fraction Decomposition
Partial-fraction decomposition is the process of starting with the simplified answer and taking it
back apart, of "decomposing" the final expression into its initial polynomial fractions. There are
four distinct cases that are explained within this handout.
Before starting, consider a rational function
f(x) =
where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions provided
that the degree of P is less than the degree of Q.
Case I: The Denominator Q(x) is a Product of Distinct Linear Factors.
Ex. f(x) = (notice that the lead power is bigger for the denominator Q than for P)
Q(x) = 2x3 + 3x2 – 2x = x(2x2 + 3x – 2) = x(2x – 1)(x + 2)
Since the denominator has three distinct factors, the partial fraction decomposition of the f(x) has
the form
We then multiply on both sides by the common denominator x(2x-1)(x+2). We then get:
Distribute and simplify:
Group like terms on the left side:
Subject: Pre-Calculus, Calculus 2, and Differential Equations
Created by: Leo Perez
Revised:2/24/2018
Partial Fraction Decomposition
The coefficients of the terms on the left should equal those of similar terms on the right.
We then solve the linear system of equations by any method which should give us:
A = 1/2 B = 1/5 C = -1/10
We can now rewrite the original irrational fraction as
Case II: Q(x) is a Product of Linear Factors, Some Repeated Factors
Ex. f(x)=
We factor the bottom to get:
We can now equate this to its partial fraction form.
From here, we can follow the first example to solve this example and the results should be:
A=1, B=2, C=-1
We can rewrite the original function like so:
Subject: Pre-Calculus, Calculus 2, and Differential Equations
Created by: Leo Perez
Revised:2/24/2018
Partial Fraction Decomposition
Case III: Q(x) Contains Irreducible Quadratic Factors, None Repeated
Ex. f(x)=
The function with the factored bottom looks like so:
We can now equate this to its partial fraction form.
From here, we can also follow example 1 to solve the problem and we should get:
A=1, B=1, C=-1
We can rewrite the original function like so:
Case IV: Q(x) Contains a Repeated Irreducible Quadratic Factor
Ex. f(x)=
The function with the factored bottom looks like:
We can now equate this to its partial fraction form.
Subject: Pre-Calculus, Calculus 2, and Differential Equations
Created by: Leo Perez
Revised:2/24/2018
Partial Fraction Decomposition
From here, we can again follow example 1 to solve the problem and the results should be:
A=1, B=0, C=-1, D=0, E=2
We can rewrite the original function like so:
Reference: The following works were referred to during the creation of this handout:
purplemath.com and Stewart Calculus: Early Transcendentals 7th edition.
