Example 1 determine the partial fraction

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Unformatted text preview: ith doing all the negative integers first. We are doing this to make a point on how we can use the fact given above to help us identify zeroes. © 2007 Paul Dawkins 268 http://tutorial.math.lamar.edu/terms.aspx College Algebra 21 3 3 -9 -9 2 -17 156 -1401 12600 = P ( -9 ) ¹ 0 -3 2 - 5 18 - 51 144 = P ( -3) ¹ 0 -1 2 - 1 4 -1 - 8 = P ( -1) ¹ 0 Now, we haven’t found a zero yet, however let’s notice that P ( -3) = 144 > 0 and P ( -1) = -8 < 0 and so by the fact above we know that there must be a zero somewhere between x = -3 and x = -1 . Now, we can also notice that x = - 3 = -1.5 is in this range and is the 2 only number in our list that is in this range and so there is a chance that this is a zero. Let’s run through synthetic division real quick to check and see if it’s a zero and to get the coefficients for Q ( x ) if it is a zero. 3 -9 2 -2 6 -6 0 2 - 3 2 1 3 So, we got a zero in the final spot which tells us that this was a zero and Q ( x ) is, Q ( x ) = 2 x3 - 2 x 2 + 6 x - 6 We now need to repeat the whole process with this polynomial. Also, unlike the previous example w...
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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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