355 3.3 Zeros of Polynomial Functions Because ƒ1x2 is a fourth-degree polynomial function, it must have four complex zeros, some of which may be repeated. Descartes’ rule of signs has indicated that exactly one of these zeros is a negative real number. • One possible combination of the zeros is one negative real zero, three positive real zeros, and no nonreal complex zeros. • Another possible combination of the zeros is one negative real zero, one positive real zero, and two nonreal complex zeros. By the conjugate zeros theorem, any possible nonreal complex zeros must occur in conjugate pairs because ƒ1x2 has real coefficients. The table below summarizes these possibilities. x 4 –1 1 –2 4 y f(x) = x4 – 6x3 + 8x2 + 2x – 1 One negative zero Three positive zeros Figure 19 Possible Number of Zeros Positive Negative Nonreal Complex 3 1 0 1 1 2 The graph of ƒ1x2 in Figure 19 verifies the correct combination of three positive real zeros with one negative real zero, as seen in the first row of the table.* S Now Try Exercise 79. NOTE Descartes’ rule of signs does not identify the multiplicity of the zeros of a function. For example, if it indicates that a function ƒ1x2 has exactly two positive real zeros, then ƒ1x2 may have two distinct positive real zeros or one positive real zero of multiplicity 2. 3.3 Exercises CONCEPT PREVIEW Determine whether each statement is true or false. If false, explain why. 1. Because x - 1 is a factor of ƒ1x2 = x6 - x4 + 2x2 - 2, we can also conclude that ƒ112 = 0. 2. Because ƒ112 = 0 for ƒ1x2 = x6 - x4 + 2x2 - 2, we can conclude that x - 1 is a factor of ƒ1x2. 3. For ƒ1x2 = 1x + 2241x - 32, the number 2 is a zero of multiplicity 4. 4. Because 2 + 3i is a zero of ƒ1x2 = x2 - 4x + 13, we can conclude that 2 - 3i is also a zero. *The authors would like to thank Mary Hill of College of DuPage for her input into Example 7.
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