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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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