So lets start with the following lets suppose that we

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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...
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