Partial Fractions

 

 

Contents

 

Partial Fractions

 

Distinct Linear Factors

1  2

Repeated Linear Factors

 

Quadratic Factors

 

Improper Fractions

 

Quiz

 

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Distinct Linear Factors
 
 
Cover-up Rule
 
There is a short cut of obtaining A and B in the previous example directly. This method is known as the "cover-up" rule.
 
Look at the identity
.
 
To find A, cover (x + 5) on the left-hand side of the identity and put in x = -5,
 
                                    A = [-(-5) - 26] / (-5-2)
                                       = 3
 
                                    B= (-2-26) / (2+5)
                                      = -4
 
The answer is the same as the one obtained by the previous method.
 
Note that the cover-up rule can only be used for distinct linear factors.
 
 
 
 Example 2:
 
Express in partial fractions:
                          
 
  Let
              .
 
    Using "cover-up" rule,
 
               To find A, substitute x = -0.5,
                      A = (-0.5 + 11) / (-0.5 + 2)
                         = 7
 
               To find B, substitute x = -2,
                      B = (-2 + 11) / [ 2 (-2) + 1]
                         = -3
 
Hence,
                
 
 
Example 3:
 
Express in partial fractions:
                            
 
  Let 
        
 
    Using "cover-up" rule, 
 
                   
                       = 1
 
                   
                       = 2
 
                   
                       = -3
 
   Therefore,