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**Unformatted text preview: **t two zeros. Since there seems to be no other rational zeros to try, we
continue with −1. Also, the shape of the crossing at x = −1 leads us to wonder if the zero
x = −1 has multiplicity 3.
−1
−1 2
4 −1 −6 −3
↓ −2 −2
3
3
2
2 −3 −3 0
↓ −2
0
3
2
0 −3 0 Success! Our quotient polynomial is now 2x2 − 3. Setting this to zero gives 2x2 − 3 = 0, or
√
3
x2 = 2 , which gives us x = ± 26 . Concerning multiplicities, based on our division, we have
that −1 has a multiplicity of at least 2. The Factor Theorem tells us our remaining zeros,
√
± 26 , each have multiplicity at least 1. However, Theorem 3.7 tells us f can have at most 4
real zeros, counting multiplicity, and so we conclude that −1 is of multiplicity exactly 2 and
√
± 26 each has multiplicity 1. (Thus, we were wrong to think that −1 had multiplicity 3.)
It is interesting to note that we could greatly improve on the graph of y = f (x) in the previous
example given to us by the calculator. For instance, from√our dete...

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