# Partial Fractions

### Written by Don Methven and generously donated to Gary and Kenny Felder's Math and Physics Help Home Page

Partial fractions: I used to find these pretty tough to solve—especially if fractions are not your strong point. Problems typically involve the splitting up of a single fraction into two or more fractions that each contain a single factor in the denominators (the bottom bit of fractions). I have written this as a quick reminder on how you use the methods to deal with them, and solved an example problem using each method. It might be helpful to print this out and follow it along by hand (math is hard to do on a screen!)

OK, let's go...
```Writing:

6		    1               1
-------------    =    -------    -    -------
x2 + 2x - 8	          x - 2	          x + 4

6
means that you have expressed -----------  in partial fractions...
x2 + 2x - 8

There are three main "methods" that allow us to do this:

(1) Substitution of strategic values

(2) Solving with coefficients

(3) Cover up method (can only be used on fractions without powers in the denominator)

Method 1

Substitution of strategic values
========================

Problem:

Express as a partial fraction

x - 1
---------------
(3x - 5)(x - 3)

========================

Solution:

First write the fraction as

x - 1               A                B
---------------   =   --------    +    -------
(3x - 5)(x - 3)       (3x - 5)         (x - 3)

Notice that I have taken the two terms that are in brackets and placed them on their own
as seperate fractions using 'A' and 'B' as numerators.

A and B are the mystery numbers we need to discover.

Next, multiply denominator (the bottom bit of the big fraction) by both sides of the equation to cancel out terms So we now have

x - 1       =   A(x - 3)   +   B(3x - 5)

In order to get rid of either the 'A' or 'B' term, we can substitute a 'strategic' value to make it equal to zero.
For example, to get rid of A(x - 3), let's make x = 3, so (3 - 3) ends up as 0. Gone.

Remember that you have to do this to all of the x's in the equation

3 - 1	      =   A(3 - 3)  +  B(3*3 - 5) you can see the 'A' component becomes zero.

3 - 1	      =   B(9 - 5)

2	      =   4B

so,	 2/4          =   B

or,	 1/2          =   B

We've discovered what B is, now we can repeat the process to discover A

Let's make x = 5/3 to get rid of the 'B' part of the equation...

so,          B(5/3 * 3) = 5

and,         B(5 - 5)   = 0

The B(3x - 5) term is gone! So we can plug this value into the entire equation to find A

5/3 - 1      =   A(5/3 - 3)

2/3          =   A * -4/3

2/3
------        =   A
-4/3

-1/2         =   A

We now have A and B, and the answer to our problem.

x - 1               -1              1
---------------   =   ---------   +   --------
(3x - 5)(x - 3)       2(3x - 5)       2(x - 3)

#########################################################################################

Method 2

Solving by coefficients
========================
```
Usually people use this method in conjunction with another method (e.g. substitution of values), however there is nothing wrong with using this method on its own if you prefer it.
```========================

Problem:

Express as a partial fraction

x - 1
---------------
(3x - 5)(x - 3)

========================

Solution:

First write the fraction as

x - 1               A                B
---------------   =   --------    +    -------
(3x - 5)(x - 3)       (3x - 5)         (x - 3)

The first step (just as in the first method) is to multiply the denominator by both sides

x - 1        =   A(x - 3)   +   B(3x - 5)

It may help if the A and B parts are expanded but this step can usually be missed.

x - 1        =   A*x - A*3    +     B*3*x - 5B

x - 1        =   Ax - 3A    +    3Bx - 5B

now look at this and try to equate coefficients for 'x' Ok, we've now done the first step, let's do the same again but with something else (like x2 or
constants (the constants are the actual number's, i.e not x's which can be anything).

Coefficient for constants

-1	       =  -3A  -  5B

we now have a simultanious equation...

1       =     A  +  3B				--- equation (1)

-1        =  -3A   + (-5B)				--- equation (2)

We can easily solve this

First, multiply equation (1) by -3

-3        = -3A   + (-9B)				--- equation (1a)

Next, subtract equation (1a) from equation (2)

2	       =  4B

2/4       =   B

0.5       =   B

Then we can plug this value into equation (1) to get A 1       =   A   +   3 * 0.5

1   -   3 * 0.5       =   A

-0.5		    =	A

So we now know A and B

x - 1               -0.5           0.5
---------------   =   ---------   +   --------
(3x - 5)(x - 3)       (3x - 5)        (x - 3)

Finally, the "0.5's" can be written with "1/2's" instead

x - 1               -1              1
---------------   =   ---------   +   --------
(3x - 5)(x - 3)       2(3x - 5)       2(x - 3)

#########################################################################################

Method 3

Cover up method
========================
```
This method is (in my opinion) the easiest of all the three methods, but it can be misleading if you follow it like a cookbook recipe without knowing how it really works. Be wary that it cannot be used with "non-linear" fractions (explained after this section).
```========================

Problem:

Express as a partial fraction

x - 1
---------------
(3x - 5)(x - 3)

========================

Solution:

Delete first term in the denominator
3x - 5
3x - 5 = 0
3x     = 5
x     = 5/3

Plug 5/3 as 'x' into what is left of the fraction

5/3 - 1              2/3            -1
-------------   =   ------------   =   ---
5/3 - 3             -4/3             2

We now have the answer for the (3x - 5) expression in the partial fraction.

Let's repeat for other expression

First, delete second term

x - 3
x - 3 = 0
x     = 3

Plug in 3 as 'x' for what is left of the fraction

3 - 1           2         1
-----------   =   ---   =   ---
3*3 - 5          4         2

This is the answer for the (x - 3) expression of the partial fraction

so,

x - 1               -1              1
---------------   =   ---------   +   --------
(3x - 5)(x - 3)       2(3x - 5)       2(x - 3)

Done.
#########################################################################################

Strange exception 1 (repeated-linear fractions)

Where a fraction has a repeated linear part,  e.g. (1 + x)2 the format will look something like

numerator                     A                 B               C
-------------------    =    ------------   +    ---------   +   ---------
expression*(1 + x)2	     expression          (1 + x)         (1 + x)2

```
Notice that becuase (1 + x)2 is in the denominator it also has an (1 + x) not squared to go with it.
```Example
========================

Problem:

Express as a partial fraction

1
---------------
(x - 3)(x + 1)2

========================

Solution:

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

Multiply both sides by the denominator

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

Expand out (in this case its only the 'A' term)

1	      =   A(x + 1)(x + 1)   +   B(x - 3)(x + 1)   +   C(x - 3)

Find A by making x = 3 (this is an 'substitution of strategic values' part)

1	      =   A(3 + 1)(3 + 1)

1          =   A*16

1/16         =   A

Lets also get C by making x = -1

1          =   C(-1 - 3)

1          =   C*-4

-1/4	      =   C

We now know A and C, so can easily get B. Lets make x = 1, so nothing gets cancelled out.

In full,   1	      =   A(1 + 1)(1 + 1)   +   B(1 - 3)(1 + 1)   +   C(1 - 3)

1          =   A*4 + B*-4 + C*-2

1          =   (1/16)*4 + B*-4 + (-1/4)*-2

1          =   1/4 + -4B + 2/4(-1/4)*-2

1	      =   3/4 - 4B

1 - 3/4    = -4B

1/4        = -4B

-1/16       =   B

Solved.

1		       1                 1                 1
---------------   =   -----------   -   -----------   -   ------------
(x - 3)(x + 1)2        16(x - 3)         16(x + 1)          4(x + 1)2

#########################################################################################

Strange exception 2 (quadratics in the denominator)
```
If you have got a quadratic in the denominator i.e. x2 + 3x + 2 the format of the partial fraction should look like this
```			      A              Bx + C
fraction       =  ---------   +   ------------

Example
========================

Problem:

Express as a partial fraction

x - 1
--------------------
(x + 3)(x2 + 3x + 2)

========================

Solution:

x - 1		   A		   Bx + C
--------------------   =   ---------   +   -------------
(x + 3)(x2 + 3x + 2)	         x + 3          x2 + 3x + 2

First multiply by denominator

x - 1		   =  A(x2 + 3x + 2)   +   (Bx + C)(x + 3)

Using our substitution of strategic values method, we can make x = -3 (at this point
be careful that your selected value doesn't also make the quadratic equal zero as well!)

-3 - 1		  =   A((-3)2 + 3*(-3) + 2)

-4		  =   A(9 - 9 + 2)

-4		  =   2A

-2		  =   A

We can use the coefficient method, looking at the coefficients of x2

0		  =   A + B

0		  =  -2 + B

2		  =   B

Similarly, looking at the coefficients of x

1		  =   3A + 3B + C

1		  =   3*(-2) + 3*(2) + C

1		  =   C

Done.

x - 1	          -2            2x  +  1
--------------------   =   ---------   +   ---------------
(x + 3)(x2 + 3x + 2)        (x + 3)         x2 + 3x + 2

```
Hope all of that helped! Gary and Kenny Felder's Math and Physics Help Home Page
`www.felderbooks.com/papers`