Heaviside’s Cover-up Method
Example 1 (Only for linear terms in the denominator)
8x 2  3 x  21
8 x 2  3 x  21

x3  7 x  6
 x  2  x  3 x  1
Now, we set this factored function equal to a sum of smaller fractions, each of which has
one of the factored terms for a denominator.
8x 2  3x  21
A
B
C



3
x  7x  6
x  2 x  3 x 1
To find A
(Think: Multiply both sides of the equation by (x+2) and then let x = -2, all of the terms
on the right-hand side will go to zero except for A and the (x+2) will cancel on the left8x 2  3x  21
A
B
C
side.  x  2 
  x  2
  x  2
  x  2
x 3
x 1
x2
 x  2  x  3 x  1
Do:
8 x 2  3x  21
 x  2   x  3 x  1 x2
Calculate: B
8 x 2  3x  21

 x  2  x  3 x  1 x3
Calculate C
8 x 2  3x  21

 x  2  x  3 x  1 x1
8  2   3  2   21
2

 2  3 2  1
1