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Home > Algebra calculators > Partial Fraction decomposition example
Partial Fraction decomposition example ( Enter your problem )
( Enter your problem )
  1. Example `(5x-4)/(x^2-x-2)`
  2. Example `(3x)/(x^2+2x+1)`
  3. Example `(x-3)/(x^3+2x^2+x)`
  4. Example `(x^2+15)/((x+3)^2(x^2+3))`
  5. Example `(2x^3)/((x+1)(x-1))`
  6. Example `(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)`
4. Example `(x^2+15)/((x+3)^2(x^2+3))`
(Previous example)		6. Example `(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)`
(Next example)

5. Example `(2x^3)/((x+1)(x-1))`


Partial Fraction `(2x^3)/((x+1)(x-1))`

Solution:
The numerator `2x^3` is of degree 3 and the denominator `x^2-1` is of degree 2. So first do the long division

long division of `2x^3` and `x^2-1`


Final Solution
 ```2x``+``0`  
`color{blue}{x^2-1}``` `2x^3``+` `0x^2``+` `0x``+` `0`  
 ``−`2x^3``-`+`2x` `2x xx (color{blue}{x^2-1})`
 `` `0x^2``+` `2x``+` `0`  
 ``−`0x^2``-`+`0` `color{green}{0} xx (color{blue}{x^2-1})`
 `` `2x`  

Final answer `= "Quotient" + (color{Magenta}{"Remainder"})/(color{blue}{"Divisor"})`.
`:.` Final answer = `2x+0 + (color{Magenta}{2x})/(color{blue}{x^2-1})`
 
Here, Divisor = `x^2-1`
Dividend = `2x^3`
Quotient = `2x+0`
Remainder = `2x`


Step by step division solution

Step - 1 :
1. Divide the first term of the dividend by the first term of the divisor : `(2x^3)/(x^2)=color{green}{2x}`

2. Write down the calculated result `color{green}{2x}` in the upper part of the table.

3. Multiply it by the divisor `color{green}{2x} xx (color{blue}{x^2-1})=color{red}{2x^3-2x}`

4. Subtract this result from the dividend
`(2x^3+0x^2+0x+0)-(color{red}{2x^3-2x})=color{Magenta}{0x^2+2x+0}`

 ```2x`  
`color{blue}{x^2-1}``` `2x^3``+` `0x^2``+` `0x``+` `0`  
 ``−`2x^3``-`+`2x` `color{green}{2x} xx (color{blue}{x^2-1})`
 `` `0x^2``+` `2x``+` `0`  


Step - 2 :
1. Divide the first term of the dividend by the first term of the divisor : `(0x^2)/(x^2)=color{green}{0}`

2. Write down the calculated result `color{green}{0}` in the upper part of the table.

3. Multiply it by the divisor `color{green}{0} xx (color{blue}{x^2-1})=color{red}{0x^2-0}`

4. Subtract this result from the remainder
`(0x^2+2x+0)-(color{red}{0x^2-0})=color{Magenta}{2x}`

 ```2x``+``0`  
`color{blue}{x^2-1}``` `2x^3``+` `0x^2``+` `0x``+` `0`  
 ``−`2x^3``-`+`2x` `2x xx (color{blue}{x^2-1})`
 `` `0x^2``+` `2x``+` `0`  
 ``−`0x^2``-`+`0` `color{green}{0} xx (color{blue}{x^2-1})`
 `` `2x`  


The long division gives
`(2x^3)/(x^2-1) = 2x + (2x)/(x^2-1)`

Now find partial fraction of `(2x)/(x^2-1)`


1. Factors the denominator
`(2x)/(x^2-1)=(2x)/((x-1)(x+1))`

2. Partial fraction for each factors
`:. (2x)/((x-1)(x+1))=A/(x-1)+B/(x+1)`

3. Multiply through by the common denominator of `(x-1)(x+1)`

`:. 2x=A xx (x+1)+B xx (x-1)`

`:. 2x=Ax+A+Bx-B`

4. Group the `x`-terms and the constant terms

`:. 2x=(A+B)x+(A-B)`

5. Coefficients of the two polynomials must be equal, so we get equations
`A+B=2`

`A-B=0`

Solution of equations using Elimination method

Total Equations are `2`

`a+b=2 -> (1)`

`a-b=0 -> (2)`


Select the equations `(1)` and `(2)`, and eliminate the variable `a`.

`a+b=2`` xx 1->````a``+``b``=``2```
−
`a-b=0`` xx 1->````a``-``b``=``0```
```2b``=``2`` -> (3)`


Now use back substitution method
From (3)
`2b=2`

`=>b=(2)/(2)=1`

From (1)
`a+b=2`

`=>a+(1)=2`

`=>a+1=2`

`=>a=2-1=1`

Solution using back substitution method.
`a=1,b=1`



After solving these equations, we get
`a=1,b=1`

Substitute these values in the original fraction
`((2x^3))/((x-1)(x+1))=2x+(1)/(x-1)+(1)/(x+1)`



This material is intended as a summary. Use your textbook for detail explanation.
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4. Example `(x^2+15)/((x+3)^2(x^2+3))`
(Previous example)		6. Example `(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)`
(Next example)


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