# Partial Fraction Calculator + Online Solver With Free Steps

A **Partial Fraction Calculator** is used to solve Partial Fraction problems. This calculator results in two constituent fractions which make up the original fraction in our problems, and the process used is **Partial Fraction Expansion**.

## What Is a Partial Fraction Calculator?

**The Partial Fraction Calculator is an online calculator which is designed to solve a polynomial fraction into its constituent fractions.**

This calculator works by using the method of **Partial Fraction Expansion**.

We will look into it more as we move forward.

## How to Use the Partial Fraction Calculator?

To use the **Partial Fraction Calculator**, you must enter the numerator and the denominator into the input boxes and press the Submit button. Now, a step-by-step guide to using this** Calculator **can be seen here:

### Step 1

Enter the numerator and the denominator in their corresponding input boxes.

### Step 2

Press the “Submit” button and it will generate the solution to your problem.

### Step 3

If you want to keep using the calculator enter new inputs and get newer results. There is no limit to the number of times you can use this calculator.

## How Does the Partial Fraction Calculator Work?

The **Partial Fraction Calculator** works by solving the **Polynomial Fraction** provided to it into its constituent fractions by using the method of partial fractions. It is also referred to as the **Partial Fraction Expansion**, and we shall go deeper into this method further in this article.

Now, let’s look at the polynomials which make up a fraction.

### Polynomials

**Polynomials** represent the class of **Mathematical Functions** that are expressed in a certain format, this may include algebraic, exponential, major mathematical operations, etc.

Now, two fractional polynomials when added together can lead to another** Polynomial**. And this process is called the LCM or also known as the **Least Common Multiple**. And now we shall look into this method underneath.

### Least Common Multiple

Now, **Least Common Multiple** is a very common method for solving fractions adding together. It is globally known as **LCM**, and its working can be seen as follows.

Here, we will assume a couple of polynomial fractions:

\[ \frac {p} {q} + \frac {r} {s} \]

To solve this problem, we must multiply the** Denominator** of each fraction by the numerator of the other, and also multiply them both to each other to create a new** Denominator**.

This can be seen in action as follows:

\[ \frac{ p \times s } { q \times s } + \frac { r \times q } { s \times q } = \frac { ( p \times s ) + ( r \times q ) } { q \times s } \]

One may wonder that this method is not being used in the **Ultimate Solution**, but it is indeed important to know the workings of this method. Given that the method we are looking into, namely the **Partial Fraction Expansion** method is the opposite of this **Mathematical Process**.

### Partial Fractions

**A partial Fraction** is a method for converting a fraction into its constituent polynomials that would have been added together to make this fraction using the **LCM method**. Now, we can delve deeper into how this method works and solves a **Fraction** into two fractions.

Let there be a polynomial fraction, and it is expressed as follows:

\[ f (x) = \frac {p(x)} {q_1(x) q_2(x)} \]

Here, we will assume numerators for two fractions which would make this fraction and name them $A$ and $B$. This is done here:

\[ f (x) = \frac {p(x)} { q_1(x) q_2(x)} = \frac {A} {q_1(x)} + \frac {B} {q_2(x)} \]

Now, we shall take the denominator from the original fraction and multiply and divide it on both sides of the equation. This can be seen here:

\[ p(x) = \frac {A} {q_1(x)} \times ( q_1(x) q_2(x) ) + \frac {B} {q_2(x)} \times ( q_1(x) q_2(x) ) \]

\[ p(x) = A \times q_2(x) + B \times q_1(x) \]

At this point, we take the expressions $q_1(x)$ and $q_2(x)$ and solve them separately by putting them against zero. This produces two results, one in which the term containing $q_1(x)$ turns to zero, and another where $q_2(x)$ turns to zero. Thus, we get our values of $A$ and $B$.

\[ Where, \phantom {()} q_1(x) = 0, \phantom {()} p(x) = A \times q_2(x), \phantom {()} \frac { p(x) }{ q_2(x) } = A \]

Similarly,

\[ Where, \phantom {()} q_2(x) = 0, \phantom {()} p(x) = B \times q_1(x), \phantom {()} \frac { p(x) }{ q_1(x) } = B \]

Here we mainly compare the **Variables** to get our results. Thus, we get the solution to our partial fractions problem.

## Solved Examples

Now let’s look at some examples to understand the concepts better.

### Example 1

Consider the polynomial fraction:

\[ \frac { 5x – 4 } { x^2 – x – 2 } \]

Solve the fraction using partial fractions.

### Solution

First, we spilt the denominator into two parts based on factorization. It can be seen done here:

\[ \frac { 5x – 4 } { x^2 – x – 2 } = \frac { 5x – 4 } { ( x – 2 ) ( x + 1 ) } \]

Now, let’s make the numerator split into $A$, and $B$. And this is done here:

\[ \frac { 5x – 4 } { ( x – 2 ) ( x + 1 ) } = \frac { A } { ( x – 2 ) } + \frac { B } { ( x + 1 ) } \]

Here, we shall multiply and divide the denominator on both sides.

\[ 5x – 4 = A ( x + 1 ) + B ( x – 2 ) \]

Then we have to place in the value of $ x + 1 = 0 $, which results in $ x = -1 $.

\[ 5( -1) – 4 = A ( -1 + 1 ) + B ( -1 – 2 ) \]

\[ – 5 – 4 = A ( 0 ) + B ( – 3 ) \]

\[ – 9 = -3 B \]

\[ B = 3 \]

Now, we repeat the process with $ x – 2 = 0 $, which results in $ x = 2 $.

\[ 5( 2 ) – 4 = A ( 2 + 1 ) + B ( 2 – 2 ) \]

\[ 10 – 4 = A ( 3 ) + B ( 0 ) \]

\[ 6 = 3 A \]

\[ A = 2 \]

Finally, we get:

\[ \frac { 5x – 4 } { ( x – 2 ) ( x + 1 ) } = \frac { A } { ( x – 2 ) } + \frac { B } { ( x + 1 ) } = \frac { 2 } { ( x – 2 ) } + \frac { 3 } { ( x + 1 ) } \]

We have our constituent fractions.

### Example 2

Consider the fraction:

\[ \frac { x^2 + 15 } { ( x + 3 )^2 ( x^2 + 3 ) } \]

Calculate the constituent fractions for this fraction using the **Partial Fraction Expansion**.

### Solution

First, we set it up in the partial fraction form:

\[ \frac { x^2 + 15 } { ( x + 3 )^2 ( x^2 + 3 ) } = \frac{A}{ ( x + 3 ) } + \frac{B}{ ( x + 3 )^2 } + \frac{Cx+D}{ ( x^2 + 3 ) } \]

Now, solve for denominator:

\[ x^2 + 15 = A ( x + 3 ) ( x^2 + 3 ) + B ( x^2 + 3 ) + (Cx + D) ( x + 3 )^2 \]

Now solve for $ x = -3 $, which can be seen here:

\[ (-3)^2 + 15 = A ( -3 + 3 ) ( (-3)^2 + 3 ) + B ( (-3)^2 + 3 ) + (C(-3) + D) ( -3 + 3 )^2 \]

\[ 9 + 15 = 0 + B ( 9 + 3 ) + 0 \]

\[ 24 = B ( 12 ) \]

\[ B = 2 \]

Now we move forward by placing the value of $B$ in the first equation, and then comparing the variables on both ends.

\[ x^2 + 15 = A ( x + 3 ) ( x^2 + 3 ) + 2 ( x^2 + 3 ) + (Cx + D) ( x + 3 )^2 \]

Then we get:

\[ x^2+15 = x^3(A + C) + x^2(3A + 6C + D + 2) + x(3A + 9C + 6D) + (9A + 6 + 9D) \]

Hence, the comparison leads to:

\[x^3 : 0 = A + C\]

\[x^2 : 1 = 3A + 6C + D + 2\]

\[x : 0 = 3A + 9C + 6D\]

\[Constants : 15 = 9A + 6 + 9D \]

\[ A = \frac{1}{2}, \phantom{()} B = 2, \phantom{()} C = \frac{-1}{2} \phantom{()} D = \frac{1}{2} \]

Thus, the partial fraction solution is:

\[ \frac { x^2 + 15 } { ( x + 3 )^2 ( x^2 + 3 ) } = \frac{\frac{1}{2}, }{ ( x + 3 ) } + \frac{2}{ ( x + 3 )^2 } + \frac{(\frac{-1}{2})x+\frac{1}{2} }{ ( x^2 + 3 ) } \]