 # 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:

$\frac{ p x s } { q x s } + \frac { r x q } { s x q } = \frac { ( p x s ) + ( r x q ) } { q x s }$

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 that 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 . q2(x) + B . q1(x)

At this point, we take the expressions q1(x) and q2(x) and solve them separately by putting them against zero. This produces two results, one in which the term containing q1(x) turns to zero, and another where q2(x) turns to zero. Thus, we get our values of A and B.

Where, q1(x) = 0, p(x) = A x q2(x),  $\frac { p(x) }{ q_2(x) }$ = A

Similarly,

Where, q2(x) = 0, p(x) = B x q1(x),  $\frac { p(x) }{ q_1(x) }$ = 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 the 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:

$x^2+15 = x^3(A + C) + x^2(3A + 6C + D + 2) + x(3A + 9C + 6D) + (9A + 6 + 9D)$

$x^3$ : 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:

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