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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.
 Spring '12
 MrVinh

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