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Unformatted text preview: eck them. Let’s again start with the integers and see what we get. 2 -2 6 -6
1 2 0 6 0 = P (1) = 0 ! !
So, x = 1 is a zero of Q ( x ) and we can now write Q ( x ) as, Q ( x ) = 2 x3 - 2 x 2 + 6 x - 6 = ( x - 1) ( 2 x 2 + 6 ) and as with the previous example we can solve the quadratic by other means. 2x2 + 6 = 0
x 2 = -3
x = ± 3i
So, in this case we get a couple of complex zeroes. That can happen.
Here is a complete list of all the zeroes for P ( x ) and note that they all have multiplicity of one. 3
x = - , x = 1, x = - 3 i, x = 3 i
So, as you can see this is a fairly lengthy process and we only did the work for two 4th degree
polynomials. The larger the degree the longer and more complicated the process. With that
being said, however, it is sometimes a process that we’ve got to go through to get zeroes of a
polynomial. © 2007 Paul Dawkins 270 http://tutorial.math.lamar.edu/terms.aspx College Algebra Partial Fractions
This section doesn’t really have a lot to do with the rest of this chapte...
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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