Also unlike the previous example we cant just reuse

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Unformatted text preview: iting down the third row and since we will be going through the process multiple time we put all the rows into a table. Here is the first synthetic division table for this problem. 1 -7 17 -17 6 -1 1 -8 25 -42 48 = P ( -1) ¹ 0 1 1 -6 11 -6 0 = P (1) = 0 ! ! So, we found a zero. Before getting into that let’s recap the computations here to make sure you can do them. © 2007 Paul Dawkins 266 http://tutorial.math.lamar.edu/terms.aspx College Algebra The top row is the coefficients from the polynomial and the first column is the numbers that we’re evaluating the polynomial at. Each row (after the first) is the third row from the synthetic division process. Let’s quickly look at the first couple of numbers in the second row. The number in the second column is the first coefficient dropped down. The number in the third column is then found by multiplying the -1 by 1 and adding to the -7. This gives the -8. For the fourth number is then -1 times -8 added onto 17. This is 25, etc. You can do regul...
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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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