Okay back to the problem we now know that x 1 is a

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Unformatted text preview: ice however, that the four fractions all reduce down to two possible numbers. This will always happen with these kinds of fractions. What we’ll do from now on is form the fraction, do any simplification of the numbers, ignoring the ± , and then drop one of the ± . So, the list possible rational zeroes for this polynomial is, © 2007 Paul Dawkins 264 http://tutorial.math.lamar.edu/terms.aspx College Algebra ±1 = ±1 ±1 ±2 = ±2 ±1 ±3 = ±3 ±1 ±6 = ±6 ±1 So, it looks there are only 8 possible rational zeroes and in this case they are all integers. Note as well that any rational zeroes of this polynomial WILL be somewhere in this list, although we haven’t found them yet. [Return to Problems] (b) P ( x ) = 2 x 4 + x 3 + 3 x 2 + 3 x - 9 We’ll not put quite as much detail into this one. First get a list of all factors of -9 and 2. Note that the minus sign on the 9 isn’t really all that important since we will still get a ± on each of the factors. -9 : 2: ±1, ± 3, ± 9 ±1, ± 2 Now, the factors of -9 are all the possible numerato...
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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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