Note however that this is where the work often gets

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Unformatted text preview: ays work so we’ll be using that here. Here we are going to make use of the fact that this equation must be true regardless of the x that we plug in. So let’s pick an x, plug it in and see what happens. For no apparent reason let’s try plugging in x = 3 . Doing this gives, 8 ( 3) - 42 = A ( 3 - 3) + B ( 3 + 6 ) -18 = 9 B -2 = B Can you see why we choose this number? By choosing x = 3 we got the term involving A to drop out and we were left with a simple equation that we can solve for B. Now, we could also choose x = -6 for exactly the same reason. Here is what happens if we use this value of x. 8 ( -6 ) - 42 = A ( -6 - 3) + B ( -6 + 6 ) -90 = -9 A 10 = A So, by correctly picking x we were able to quickly and easily get the values of A and B. So, all that we need to do at this point is plug them in to finish the problem. Here is the partial fraction decomposition for this part. 8 x - 42 10 -2 10 2 = + = x + 3 x - 18 x + 6 x - 3 x + 6 x - 3 2 Notice that we moved the minus sign on the second term down to make the addition a subtraction. We will...
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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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