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Example 3.3.6. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3.
1. Find all real zeros of f and their multiplicities.
2. Sketch the graph of y = f (x).
1. We know from Cauchy’s Bound that all of the real zeros lie in the interval [−4, 4] and that
our possible rational zeros are ± 1 , ± 1, , ± 3 , and ± 3. Descartes’ Rule of Signs guarantees
us at least one negative real zero and exactly one positive real zero, counting multiplicity. We
try our positive rational zeros, starting with the smallest, 2 . Since the remainder isn’t zero,
we know 2 isn’t a zero. Sadly, the ﬁnal line in the division tableau has both positive and
negative numbers, so 1 is not an upper bound. The only information we get from this division
is courtesy of the Remainder Theorem which tells us f 2 = − 45 so the point 2 , − 45 is
on the graph of f . We continue to our next possible zero, 1. As before, the only information
we can glean from this is that (1, −4) is on the graph of f . When we try our next possible
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