However you will find that once you get past the

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Unformatted text preview: 9 x + 25 ( x + 3) 2 = A B + x + 3 ( x + 3)2 Adding the two terms together gives, 9 x + 25 ( x + 3) 2 = A ( x + 3) + B ( x + 3) 2 Notice that in this case the second term already had the LCD under it and so we didn’t need to add anything in that time. Setting the numerators equal gives, 9 x + 25 = A ( x + 3) + B Now, again, we can get B for free by picking x = -3 . © 2007 Paul Dawkins 275 http://tutorial.math.lamar.edu/terms.aspx College Algebra 9 ( -3) + 25 = A ( -3 + 3) + B -2 = B To find A we will do the same thing that we did in the previous part. We’ll use x = 0 and the fact that we know what B is. 25 = A ( 3) - 2 27 = 3 A 9= A In this case, notice that the constant in the numerator of the first isn’t zero as it was in the previous part. Here is the partial fraction decomposition for this part. 9 x + 25 ( x + 3) 2 = 9 2 x + 3 ( x + 3)2 [Return to Problems] Now, we need to do a set of examples with quadratic factors. Note however, that this is where the work often gets fairly messy and in fact we haven’t covered the material yet that will allow us to work many of these problems. We can work some simple examples however, so let’s do that....
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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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