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Unformatted text preview: 9  9x
1
5
=
2x + 7 x  4 2x 1 x + 4
2 [Return to Problems] (c) 4x2 ( x  1)( x  2 ) 2 In this case the denominator has already been factored for us. Notice as well that we’ve now got
a linear factor to a power. So, recall from our table that this means we will get 2 terms in the
partial fraction decomposition from this factor. Here is the form of the partial fraction
decomposition for this expression. 4 x2 ( x  1)( x  2 ) 2 = A
B
C
+
+
x  1 x  2 ( x  2 )2 Now, remember that the LCD is just the denominator of the original expression so in this case
we’ve got ( x  1)( x  2 ) . Adding the three terms back up gives us,
2 ( x  1)( x  2 ) A ( x  2 ) + B ( x  1)( x  2 ) + C ( x  1)
2 4 x2
2 = ( x  1)( x  2 ) 2 Remember that we just need to add in the factors that are missing to each term.
Now set the numerators equal. 4 x 2 = A ( x  2 ) + B ( x  1)( x  2 ) + C ( x  1)
2 In this case we’ve got a slightly different situation from the previous two parts. Let’s start by
picking a couple of values of x and seeing what we get since there are two that should jump right
© 2007 Paul Dawkins 274 http://tutoria...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
 Spring '12
 MrVinh

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