But how do we know if a general polynomial has any

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Unformatted text preview: ents of the quotient polynomial and the remainder are nonnegative. (Note the leading coefficient of q is the same as f and so q (x) is not the zero polynomial.) If b > c, then f (b) = (b − c)q (b) + r, where (b − c) and q (b) are both positive and r ≥ 0. Hence f (b) > 0 which shows b cannot be a zero of f . Thus no real number b > c can be a zero of f , as required. A similar argument proves f (b) < 0 if all the numbers in the final line of the synthetic division tableau are non-positive. To prove the lower bound part of the theorem, we note that a lower bound for the negative real zeros of f (x) is an upper bound for the positive real zeros of f (−x). Applying the upper bound portion to f (−x) gives the result. (Do you see where the alternating signs come in?) With the additional mathematical machinery of Descartes’ Rule of Signs and the Upper and Lower Bounds Theorem, we can find the real zeros of f (x) = 2x4 + 4x3 − x2 − 6x − 3 without the use of a graphin...
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