Unformatted text preview: re there any that can’t be rational zeroes of the original polynomial. If
there are some, throw them out as we will already know that they won’t work. So, a reduced list
of numbers to try here is, 1, ± 2, ± 3, ± 6
Note that we do need to include x = 1 in the list since it is possible for a zero to occur more that
once (i.e. multiplicity greater than one).
Here is the synthetic division table for this polynomial. 1 6 11 6
1 1 5 6
0 = P (1) = 0!!
So, x = 1 is also a zero of Q ( x ) and we can now write Q ( x ) as, Q ( x ) = x 3  6 x 2 + 11x  6 = ( x  1) ( x 2  5 x + 6 ) © 2007 Paul Dawkins 267 http://tutorial.math.lamar.edu/terms.aspx College Algebra Now, technically we could continue the process with x 2  5 x + 6 , but this is a quadratic equation
and we know how to find zeroes of these without a complicated process like this so let’s just
solve this like we normally would. x 2  5 x + 6 = ( x  2 ) ( x  3) = 0 Þ x = 2, x = 3 Note that these two numbers are in the list of possible rational zeroes.
Finishing up this problem then gives the following list of zeroes fo...
View
Full
Document
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

Click to edit the document details