# 4 Types of Partial Fractions Decomposition with Examples

Decomposing a fraction into partial fractions can be considered as the reverse process of adding or subtracting two or more fractions. There are four main types of partial fractions: denominator with linear factors, denominator with irreducible quadratic factor, denominator with a repeated factor, and improper fractions.

Here, we will learn about these four types of partial fractions in detail. Also, we will look at some examples to apply the concepts.

##### ÁLGEBRA

Relevante para

Learning about the four types of partial fractions.

See types

##### ALGEBRA

Relevante for

Learning about the four types of partial fractions.

See types

## Type 1: Denominator with linear factors

Each linear factor $latex (ax+b)$ in the denominator has a corresponding partial fraction of the form

$$\frac{A}{ax+b}$$

where $latex a,~b$ and $latex A$ are constants.

For example, if we have the following fraction

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

its partial fractions have the following form:

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

The values of the constants A and B can be found by multiplying the whole expression by the denominator of the original fraction.

Then, we can use two different methods, as shown in the following examples.

### EXAMPLE 1

Decompose the following fraction into partial fractions:

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

Since we have two linear factors $latex (x+3)$ and $latex (x-2)$, their decomposition into partial fractions is:

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

Now, we multiply the whole expression by $latex (x+3)(x-2)$ to get:

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

We can use two different methods to find the values of the constants. Using the first method, we substitute $latex x=2$ to remove the constant A and find the value of B:

$$3(2)-1=A(2-2)+B(2+3)$$

$latex 5=5B$

$latex B=1$

Then, we substitute $latex x=-3$ to find the value of A:

$$3(-3)-1=A(-3-2)+B(-3+3)$$

$latex -10=-5A$

$latex A=2$

The second method consists of expanding the right-hand side of the equation to obtain:

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

$$3x-1=Ax-2A+Bx+3B$$

Comparing the coefficients of the terms that have the variable x, we have:

$latex 3=A+B$

Comparing the constant terms, we have:

$latex -1=-2A+3B$

Forming a system of equations and solving simultaneously, we have $latex A=2$ and $latex B=1$, similar to the previous method.

Therefore,

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

### EXAMPLE 2

Express the following fraction in partial fractions

$$\frac{9x^2+34x+14}{(x+2)(x^2-x-12)}$$

In this case, we can start by factoring the denominator to get only linear factors:

$$x^2-x-12=(x+3)(x-4)$$

Then, we have:

$$\frac{9x^2+34x+14}{(x+2)(x^2-x-12)}=\frac{9x^2+34x+14}{(x+2)(x+3)(x-4)}$$

Now, since we only have linear factors in the denominator, we can write as follows:

$$\frac{9x^2+34x+14}{(x+2)(x+3)(x-4)}=\frac{A}{x+2}+\frac{B}{x+3}+\frac{C}{x-4}$$

If we multiply the whole expression by $latex (x+2)(x+3)(x-4)$, we have:

$$9x^2+34x+14=A(x+3)(x-4)+B(x+2)(x-4)+C(x+2)(x+3)$$

When using $latex x=-2$, we have:

$$9(-2)^2+34(-2)+14=A(-2+3)(-2-4)+B(-2+2)(-2-4)+C(-2+2)(-2+3)$$

$latex -18=-6A$

$latex A=3$

When using $latex x=-3$, we have:

$$9(-3)^2+34(-3)+14=A(-3+3)(-3-4)+B(-3+2)(-3-4)+C(-3+2)(-3+3)$$

$latex -7=7B$

$latex B=-1$

When using $latex x=4$, we have:

$$9(4)^2+34(4)+14=A(4+3)(4-4)+B(4+2)(4-4)+C(4+2)(4+3)$$

$latex 294=42C$

$latex C=7$

Therefore, we have:

$$\frac{9x^2+34x+14}{(x+2)(x+3)(x-4)}=\frac{3}{x+2}-\frac{1}{x+3}+\frac{7}{x-4}$$

## Type 2: Denominator with an irreducible quadratic factor

Any quadratic factor of the form $latex (ax^2+bx+c)$ in the denominator has a corresponding partial fraction of the following form if it is irreducible, that is, if it cannot be factorized:

$$\frac{Ax+B}{ax^2+bx+c}$$

where $latex a,~b,~c,~A$ and $latex B$ are constants.

For example, suppose we have the following fraction

$$\frac{7x^2+2x-28}{(x-6)(x^2+3x+5)}$$

Its partial fractions have the following form:

$$\frac{7x^2+2x-28}{(x-6)(x^2+3x+5)}=\frac{A}{x-6}+\frac{Bx+C}{x^2+3x+5}$$

Then, we can find the values of A, B, and C by using the same methods seen in Type 1, as shown in the following example.

### EXAMPLE

Express the following fraction in partial fractions:

$$\frac{7x^2+2x-28}{(x-6)(x^2+3x+5)}$$

Since the quadratic expression $latex x^2+3x+5$ in the denominator cannot be factored, we can assume that the partial fractions have the following form:

$$\frac{7x^2+2x-28}{(x-6)(x^2+3x+5)}=\frac{A}{x-6}+\frac{Bx+C}{x^2+3x+5}$$

Now, we multiply the whole expression by $latex (x-6)(x^2+3x+5)$ and we have:

$$7x^2+2x-28=A(x^2+3x+5)+(Bx+C)(x-6)$$

When we substitute $latex x=6$, we can find the value of A:

$$7(6)^2+2(6)-28=A((6)^2+3(6)+5)+(B(6)+C)(6-6)$$

$latex 236=59A$

$latex A=4$

When we compare the coefficients of the terms with $latex x^2$, we have:

$latex 7=A+B$

Substituting $latex A=4$ in this equation, we obtain the value $latex B=3$.

If we now compare the constant terms, we have:

$latex -28=5A-6C$

Substituting $latex A=4$ in this equation, we obtain the value $latex C=8$.

Therefore, the partial fractions are:

$$\frac{7x^2+2x-28}{(x-6)(x^2+3x+5)}=\frac{4}{x-6}+\frac{3x+8}{x^2+3x+5}$$

## Type 3: Denominator with one repeated factor

Each repeated linear factor of the form $latex (ax+b)^2$ in the denominator has partial fractions of the following form:

$$\frac{A}{ax+b}+\frac{B}{ax+b}^2$$

where $latex a,~b,~A$ and $latex B$ are constants.

For example, suppose we have the following fraction

$$\frac{2x^2+29x-1}{(2x+1)(x-2)^2}$$

We can decompose into partial fractions by using the following form:

$$\frac{2x^2+29x-1}{(2x+1)(x-2)^2}=\frac{A}{2x+1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$

We find the values of the constants A, B and C by applying the same techniques as in the previous examples.

### EXAMPLE

Express the following fraction in partial fractions:

$$\frac{2x^2+29x-1}{(2x+1)(x-2)^2}$$

This fraction has a linear factor $latex (2x+1)$ and a repeated factor $latex (x-2)^2$. Then, its partial fractions have the form:

$$\frac{2x^2+29x-1}{(2x+1)(x-2)^2}=\frac{A}{2x+1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$

Now, we multiply the whole expression by the denominator of the original fraction $latex (2x+1)(x-2)^2$ and we have:

$$2x^2+29x-11=A(x-2)^2+B(2x+1)(x-2)+C(2x+1)$$

When using the value $latex x=-“frac{1}{2}$, we have:

$$2(-\frac{1}{2})^2+29(-\frac{1}{2})-11=A(-\frac{1}{2}-2)^2+B(2(-\frac{1}{2})+1)(-\frac{1}{2}-2)+C(2(-\frac{1}{2})+1)$$

$$-25=\frac{25}{4}A$$

$latex A=-4$

When using the value $latex x=2$, we have:

$$2(2)^2+29(2)-11=A(2-2)^2+B(2(2)+1)(2-2)+C(2(2)+1)$$

$latex 55=5C$

$latex C=11$

If we compare the coefficients of the terms with $latex x^2$, we have:

$latex 2=A+2B$

Using the value $latex A=-4$ in this equation, we find the value of $latex B=3$. Then, we have:

$$\frac{2x^2+29x-1}{(2x+1)(x-2)^2}=-\frac{4}{2x+1}+\frac{3}{x-2}+\frac{11}{(x-2)^2}$$

## Type 4: Improper fractions

An improper algebraic fraction is a fraction in which the degree of the numerator is greater than or equal to the degree of the denominator.

To simplify an improper algebraic fraction, we can divide the numerator by the denominator. When we do this, we can get the following:

• If a polynomial of degree $latex n$ is divided by a polynomial of degree $latex n$, the quotient is a constant.
• If a polynomial of degree $latex n$ is divided by a polynomial of degree $latex m$, where $latex m<n$, the quotient is a polynomial of degree $latex (n-m)$.

For example, in the following fraction, the degree of the numerator is 2 and the degree of the denominator is also 2:

$$\frac{5x^2-71}{(x+5)(x-4)}$$

Then, the quotient is a constant (the constant A) and its partial fractions have the following form:

$$\frac{5x^2-71}{(x+5)(x-4)}=A+\frac{B}{x+5}+\frac{C}{x-4}$$

In the following fraction, the degree of the numerator is 4 and the degree of the denominator is 3:

$$\frac{3x^4+7x^3+8x^2+53x-186}{(x+4)(x^2+9)}$$

Then, the quotient is a polynomial of degree (4-3)=1 and its partial fractions have the following form:

$$\frac{3x^4+7x^3+8x^2+53x-186}{(x+4)(x^2+9)}=Ax+B+\frac{C}{x+4}+\frac{Dx+E}{x^2+9}$$

### EXAMPLE

Express the following fraction in partial fractions

$$\frac{5x^2-71}{(x+5)(x-4)}$$

The degree of the polynomial in the numerator is 2 and the degree of the polynomial in the denominator is also 2.

Then, the quotient is a constant and its partial fractions have the following form:

$$\frac{5x^2-71}{(x+5)(x-4)}=A+\frac{B}{x+5}+\frac{C}{x-4}$$

When multiplying the whole expression by $latex (x+5)(x-4)$ and we have:

$$5x^2-71=A(x+5)(x-4)+B(x-4)+C(x+5)$$

We can find the value of A easily by comparing the coefficients of the terms with $latex x^2$, and we find that $latex A=5$.

When using the value $latex x=-5$, we have:

$$5(-5)^2-71=B(-5-4)$$

$latex 54=-9B$

$latex B=-6$

When using the value $latex x=4$, we have:

$$5(4)^2-71=C(4+5)$$

$latex 9=9C$

$latex C=1$

Therefore, we have:

$$\frac{5x^2-71}{(x+5)(x-4)}=5-\frac{6}{x+5}+\frac{1}{x-4}$$