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Question 2 (1 point) State the number of complex zeros and the possible number of real and imaginary zeros for each function. \( f(x)=x^{4}-9 x^{2}+18 \) a \# of complex zeros: 4 Possible \# of real zeros: 4,2 , or 0 Possible \# of imaginary zeros: 4,2 , or 0 b \# of complex zeros: 6 Possible \# of real zeros: 4,2 , or 0 Possible \# of imaginary zeros: 4,2 , or 0 c \# of complex zeros: 8 Possible \# of real zeros: 4,2 , or 0 Possible \# of imaginary zeros: 4,2 , or 0 d \# of complex zeros: 10 Possible \# of real zeros: 4,2 , or 0 Possible \# of imaginary zeros: 4,2 , or 0

Ask by Stanley Mcdonald. in the United States
Mar 15,2025

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The function \( f(x) = x^4 - 9x^2 + 18 \) has 4 complex zeros, all of which are imaginary.

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The number of complex zeros in a polynomial is always tied to its degree, which, for the given function \( f(x) = x^4 - 9x^2 + 18 \), is 4. Thus, it has 4 complex zeros, and depending on the roots' nature, there can be various combinations of real and imaginary zeros. For possible real zeros, because it’s a quartic (degree 4) polynomial, it can have 0, 2, or 4 real zeros. Moreover, if there are any imaginary zeros, they will appear in conjugate pairs, contributing to the overall count of complex zeros. So in this case, if there are 4 real zeros, there are 0 imaginary zeros; if there are 2 real zeros, there are 2 imaginary zeros, and if there are 0 real zeros, all zeros must be imaginary.

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