 # 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 (3x - 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 3x - 2 = A(x - 1) + Bx.

A(x - 1) = Ax - A

Ax + Bx = (A + B)x

So 3x - 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 [5x2 - 1]/[x2(x + 1)] The denominator of the given fraction is
x2(x + 1).

The reduced factors of x2(x + 1) are
x, x2, and (x + 1).

Then set
(given) = A/x + B/x2 + 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:
x2(x + 1).

For each term,
multiply the missing factors
to both of the numerator and the denominator.

Compare the numerators of both sides.

Then 5x2 - 1 = Ax(x + 1) + B(x + 1) + Cx2.

Ax(x + 1) = Ax2 + Ax

B(x + 1) = Bx + B

Ax2 + Cx2 = (A + C)x2

Ax + Bx = (A + B)x

So 5x2 - 1 = (A + C)x2 + (A + B)x + B.

The coefficients of the x2 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/x2 + C/(x + 1)

So (given) = 1/x + (-1)/x2 + 4/(x + 1).

+ (-1)/x2 = -1/x2

So (given) = 1/x - 1/x2 + 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).