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Unformatted text preview: P ( x ): P (1) = (1) 32(1) 25(1) + 6 = 0. Therefore 1 is a zero of the polynomial. P (1) = (1) 32(1) 25(1) + 6 , 0. P (2) = (2) 32(2) 25(2) + 6 , 0. P (2) = (2) 32(2) 25(2) + 6 = 0. Therefore 2 is a zero of the polynomial. P (3) = (3) 32(3) 25(3) + 6 = 0. Therefore 3 is a zero of the polynomial. Since p ( x ) is a polynomial of degree it has at most three zeros. So the zeros are 1 ,2 ,3. (c) As x → + ∞ , x 3 → + ∞ = ⇒ P ( x ) → + ∞ . As x → ∞ , x 3 → ∞ = ⇒ P ( x ) → ∞ . (d) . 1...
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This note was uploaded on 09/30/2010 for the course MATH 1450 taught by Professor A during the Spring '10 term at Zhejiang University.
 Spring '10
 A
 Algebra, Trigonometry

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