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**Unformatted text preview: **in 4 sign changes, then, counting multiplicities, f has 4, 2, or
0 negative real zeros; hence, the graph of y = f (x) may not cross the negative x-axis at all. The
proof of Descartes’ Rule of Signs is a bit technical, and can be found here.
Example 3.3.5. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. Use Descartes’ Rule of Signs to determine
the possible number and location of the real zeros of f .
Solution: As noted above, the variations of sign of f (x) is 1. This means, counting multiplicities,
f has exactly 1 positive real zero. To ﬁnd the possible number of negative real zeros, we consider
f (−x) = 2(−x)4 + 4(−x)3 − (−x)2 − 6(−x) − 3 which simpliﬁes to 2x4 − 4x3 − x2 + 6x − 3. This
has 3 variations in sign, hence f has either 3 negative real zeros or 1 negative real zero, counting
multiplicities.
Cauchy’s Bound gives us a general bound on the zeros of a polynomial function. Our next result
helps us determine bounds on the real zeros of a polynomial as we synthetically divid...

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