# Partial Fraction Decomposition

How to simplify a rational expression by using the partial fraction decomposition: formula (1/*AB* form), examples, and their solutions.

## Example 1: Simplify (3*x* - 2)/[*x*(*x* - 1)]

The denominator of the given fraction is*x*(*x* - 1).

The reduced factors of *x*(*x* - 1) are*x* and (*x* - 1).

Then set

(given) = *A*/*x* + *B*/(*x* - 1).

The goal is to find *A* and *B*.

Change the denominator of each term

to the denominator of the given fraction:*x*(*x* - 1).

For each term,

multiply the missing factor

to both of the numerator and the denominator.

Compare the numerators of both sides.

Then 3*x* - 2 = *A*(*x* - 1) + *B*⋅*x*.

*A*(*x* - 1) = *A**x* - *A*

*A**x* + *B**x* = (*A* + *B*)*x*

So 3*x* - 2 = (*A* + *B*)*x* - *A*

The coefficients of the *x* terms in both sides

are the same.

So *A* + *B* = 3.

The constants in both sides

are the same.

So -*A* = -2.

This is a system of equations:*A* + *B* = 3

-*A* = -2.

Solve this system

to find *A* and *B*.

Start from the simple one:

-*A* = -2.

Multiply -1 on both sides.

Then *A* = 2.

Put [*A* = 2]

into the other equation [*A* + *B* = 3].

Then 2 + *B* = 3.

Substitution method

Move 2 to the right side.

Then *B* = 1.

*A* = 2*B* = 1

(given) = *A*/*x* + *B*/(*x* - 1)

So (given) = 2/*x* + 1/(*x* - 1).

This process is called

the partial fraction decomposition.

## Example 2: Simplify [5*x*^{2} - 1]/[*x*^{2}(*x* + 1)]

The denominator of the given fraction is*x*^{2}(*x* + 1).

The reduced factors of *x*^{2}(*x* + 1) are*x*, *x*^{2}, and (*x* + 1).

Then set

(given) = *A*/*x* + *B*/*x*^{2} + *C*/(*x* + 1).

The goal is to find *A*, *B*, and *C*.

Change the denominator of each term

to the denominator of the given fraction:*x*^{2}(*x* + 1).

For each term,

multiply the missing factors

to both of the numerator and the denominator.

Compare the numerators of both sides.

Then 5*x*^{2} - 1 = *A**x*(*x* + 1) + *B*(*x* + 1) + *C*⋅*x*^{2}.

*A**x*(*x* + 1) = *A**x*^{2} + *A**x**B*(*x* + 1) = *B**x* + *B*

*A**x*^{2} + *C**x*^{2} = (*A* + *C*)*x*^{2}*A**x* + *B**x* = (*A* + *B*)*x*

So 5*x*^{2} - 1 = (*A* + *C*)*x*^{2} + (*A* + *B*)*x* + *B*.

The coefficients of the *x*^{2} terms in both sides

are the same.

So *A* + *C* = 5.

The coefficients of the *x* terms in both sides

are the same.

So *A* + *B* = 0.

The constants in both sides

are the same.

So *B* = -1.

This is a system of equations:*A* + *C* = 5*A* + *B* = 0*B* = -1.

Solve this system

to find *A*, *B*, and *C*.

System of equations (3 variables)

Start from the simple one:*B* = -1.

Put [*B* = -1]

into [*A* + *B* = 0].

Then *A* + (-1) = 0.

Move -1 to the right side.

Then *A* = 1.

Put [*A* = 1]

into [*A* + *C* = 5].

Then 1 + *C* = 5.

Move 1 to the right side.

Then *C* = 4.

*A* = 1*B* = -1*C* = 4

(given) = *A*/*x* + *B*/*x*^{2} + *C*/(*x* + 1)

So (given) = 1/*x* + (-1)/*x*^{2} + 4/(*x* + 1).

+ (-1)/*x*^{2} = -1/*x*^{2}

So (given) = 1/*x* - 1/*x*^{2} + 4/(*x* + 1).

## Formula: 1/*AB* = 1/(*B* - *A*)(1/*A* - 1/*B*)

1/*AB* = 1/(*B* - *A*)(1/*A* - 1/*B*)

This formula is useful

when [*B* - *A*] is a constant.

## Example 3: Simplify 1/[*x*(*x* + 1)]

The difference between (*x* + 1) and *x* is 1,

which is a constant.

So use the 1/*AB* formula

to solve this example easily.*A* = *x**B* = (*x* + 1)

So 1/[*x*(*x* + 1)]

= 1/[(*x* + 1) - *x*](1/*x* - 1/(*x* + 1)].

(*x* + 1) - *x* = 1

So (given) = 1/*x* - 1/(*x* + 1).