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Stitz-Zeager_College_Algebra_e-book

# This argument gives us the following factorization

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Unformatted text preview: ero, 2 , we get that it is not a zero, and we also see that it is an upper bound on the zeros of f , since all of the numbers in the ﬁnal line of the division tableau are positive. This means there is no point trying our last possible rational zero, 3. Descartes’ Rule of Signs guaranteed us a positive real zero, and at this point we have shown this zero is irrational. Furthermore, 3 the Intermediate Value Theorem, Theorem 3.1, tells us the zero lies between 1 and 2 , since 3 f (1) < 0 and f 2 > 0. 1 2 2 4 −1 5 ↓1 2 −6 3 2 − 21 4 25 3 4 −3 − 21 8 − 45 8 1 2 4 −1 −6 ↓2 6 5 26 5 −1 −3 −1 −4 3 2 2 4 −1 −6 ↓ 3 21 57 2 4 27 19 2 33 4 −3 99 8 75 8 We now turn our attention to negative real zeros. We try the largest possible zero, − 1 . 2 Synthetic division shows us it is not a zero, nor is it a lower bound (since the numbers in 214 Polynomial Functions the ﬁnal line of the division tableau do not alternate), so we proceed to −1. This division shows −1 is...
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