Question 5 (1 point) State the number of complex zeros and the possible number of real and imaginary zeros for each function. $$ f(x)=27 x^{6}+208 x^{3}-64 $$ a \# of complex zeros: 6 Possible \# of real zeros: $$ 6,4,2 $$, Possible \# of imaginary zeros: $$ 6,4,2 $$, b \# of complex zeros: 6 Possible \# of real zeros: $$ 6,4,2 $$, or 0 Possible \# of imaginary zeros: $$ 6,4,2 $$, or c \# of complex zeros: 12 Possible \# of real zeros: $$ 6,4,2 $$, or 0 Possible \# of imaginary zeros: $$ 6,4,2 $$, o d \# of complex zeros: 10 Possible \# of real zeros: $$ 6,4,2 $$, or 0 Possible \# of imaginary zeros: $$ 6,4,2 $$, o

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We start by recalling that a polynomial of degree $$ n $$ has exactly $$ n $$ complex zeros (counting multiplicities). Since

$$
f(x)=27x^6+208x^3-64
$$

is a degree‑6 polynomial, it has exactly $$ 6 $$ complex zeros.

Because the coefficients of $$ f(x) $$ are real, any non‐real zeros must occur in conjugate pairs. This means that the number of non‑real (or imaginary) zeros is even. If we let the number of real zeros be $$ r $$, then the number of non‑real zeros is

$$
6 - r,
$$

and since $$ 6 - r $$ must be even, $$ r $$ can only be $$ 6 $$, $$ 4 $$, $$ 2 $$, or $$ 0 $$.

Thus, the function

- Has exactly $$ 6 $$ complex zeros,
- The possible number of real zeros is $$ 6 $$, $$ 4 $$, $$ 2 $$, or $$ 0 $$,
- And the possible number of non‑real (imaginary) zeros is $$ 0 $$, $$ 2 $$, $$ 4 $$, or $$ 6 $$ (since they must add up with the real zeros to $$ 6 $$).

This corresponds to answer choice (b).
The function $$ f(x) = 27x^6 + 208x^3 - 64 $$ has: - **6 complex zeros** - **Possible number of real zeros:** 6, 4, 2, or 0 - **Possible number of imaginary zeros:** 0, 2, 4, or 6 This matches answer choice (b).

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