 |
 |
 |
|
 |
|
 |
|
 |
|
 |
|
 |
 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
 |
|
|
Contents
Repeated Linear Factors
|
Repeated Linear Factors
What if one of the linear factors
in the denominator is repeated, ie.
Solution:
To express in partial fractions,
Do
note that
is not expressed as either of these: or
A and C
can be found using "cover-up" rule,
A = [ 6 (-1)2 - 21 (-1) + 9] / (-1 - 2)2
= 4
C = [ 6 (2) - 21 (2) + 9] / (2 + 1)
= -3
However, B
cannot be found using cover-up rule. Algebraic identities are
used instead.
6x2 - 21x
+ 9 = 4 (x - 2)2 + B (x + 1) (x - 2) - 3
(x + 1)
when x = 0,
4 (-2)2 - 2B - 3 = 9
B = 2
Therefore,
 
Express in partial fractions: . Let
4 + 5x - x2
= A (x + 1)2 + B (x + 1) (x - 1) +
C (x - 1)
To find A, let x
= 1,
4 + 5 - 1 = A (2)2
A = 2
To find C, let x
= -1,
4 + 5 (-1) - 1 = -2C
C = 1
To find B, let x = 0.
2 - B - 1 = 4
B = -3
Therefore,
Note that
cover-up rule can be used to find A and C.
 |
