Our last example reminds us that nding the zeros of

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Unformatted text preview: s with technology. 208 Polynomial Functions Multiplying both sides of this equation by q n , we clear the denominators to get an pn + an−1 pn−1 q + . . . + a1 pq n−1 + a0 q n = 0 Rearranging this equation, we get an pn = −an−1 pn−1 q − . . . − a1 pq n−1 − a0 q n Now, the left hand side is an integer multiple of p, and the right hand side is an integer multiple of q . (Can you see why?) This means an pn is both a multiple of p and a multiple of q . Since p and q have no common factors, an must be a multiple of q . If we rearrange the equation an pn + an−1 pn−1 q + . . . + a1 pq n−1 + a0 q n = 0 as a0 q n = −an pn − an−1 pn−1 q − . . . − a1 pq n−1 we can play the same game and conclude a0 is a multiple of p, and we have the result. Example 3.3.2. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. Use the Rational Zeros Theorem to list all possible rational zeros of f . Solution. To generate a complete list of rational zeros, we need to take each of the factors of const...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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