This is not a problem and in fact when this happens

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Unformatted text preview: l.math.lamar.edu/terms.aspx College Algebra out at us as being particularly useful. x = 1: 4 = A ( -1) x = 2: 16 = C (1) 2 Þ A=4 Þ C = 16 So, we can get A and C in the same manner that we’ve been using to this point. However, there is no value of x that will allow us to eliminate the first and third term leaving only the middle term that we can use to solve for B. While this may appear to be a problem it actually isn’t. At this point we know two of the three constants. So all we need to do is chose any other value of x that would be easy to work with ( x = 0 seems particularly useful here), plug that in along with the values of A and C and we’ll get a simple equation that we can solve for B. Here is that work. 4 ( 0 ) = ( 4 ) ( -2 ) + B ( -1) ( -2 ) + 16 ( -1) 2 2 0 = 16 + 2 B - 16 0 = 2B 0=B In this case we got B = 0 this will happen on occasion, but do not expect it to happen in all cases. Here is the partial fraction decomposition for this part. 4 x2 ( x - 1)( x - 2 ) 2 = 4 16 + x - 1 ( x - 2 )2 [Return to Problems] (d) 9 x + 25 ( x + 3) 2 Again, the denominator has already been factored for us. In this case the form of the partial fraction decomposition is,...
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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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