Alg_Complete

# Here is the first synthetic division table for this

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Unformatted text preview: be a factor of 1. Also, as we saw in the previous example we can’t forget negative factors. So, the first thing to do is actually to list all possible factors of 1 and 6. Here they are. 6: 1: ±1, ± 2, ± 3, ± 6 ±1 Now, to get a list of possible rational zeroes of the polynomial all we need to do is write down all possible fractions that we can form from these numbers where the numerators must be factors of 6 and the denominators must be factors of 1. This is actually easier than it might at first appear to be. There is a very simple shorthanded way of doing this. Let’s go through the first one in detail then we’ll do the rest quicker. First, take the first factor from the numerator list, including the ± , and divide this by the first factor (okay, only factor in this case) from the denominator list, again including the ± . Doing this gives, ±1 ±1 This looks like a mess, but it isn’t too bad. There are four fractions here. They are, +1 =1 +1 +1 = -1 -1 -1 = -1 +1 -1 =1 -1 Not...
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