Alg_Complete

# Alg_Complete

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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.

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