Question 3 (1 point) State the number of complex zeros and the possible number of real and imaginary zeros for each function \( f(x)=5 x^{5}+36 x^{3}+7 x \) a \( \quad \) \# of complex zeros: 7 Possible \# of real zeros: 5,3 , or 1 Possible \# of imaginary zeros: 4,2 , or b \( \quad \) \# of complex zeros: 9 Possible \# of real zeros: 5,3 , or 1 Possible \# of imaginary zeros: 4,2 , or c \( \quad \) \# of complex zeros: 5 Possible \# of real zeros: 5,3 , or 1 Possible \# of imaginary zeros: 4, 2, or 0 d \( \quad \) \# of complex zeros: 5 Possible \# of real zeros: 3 , or 0 Possible \# of imaginary zeros: 4, 2, or 0

When analyzing polynomial functions, the Fundamental Theorem of Algebra tells us that a polynomial of degree \( n \) must have exactly \( n \) roots (counting multiplicities), which can be real or complex. For the function \( f(x) = 5x^5 + 36x^3 + 7x \), we have a degree of 5, indicating there are 5 complex zeros in total. From the nature of polynomials, the possible number of real zeros can indeed be odd, hence we can have 1, 3, or all 5 being real, while the rest can be complex or imaginary. To avoid common pitfalls, remember that imaginary zeros always come in pairs, so if there are any imaginary zeros, they will affect the count of real zeros too. For instance, if you have 1 real zero, you must have 4 complex zeros (which could be two pairs of imaginary zeros) while ensuring the total always adds up to 5 for the function’s degree!