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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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