Answer: C) 3 real zeros & 2 imaginary zeros
Step-by-step explanation:
Since the function has a degree of 5, then there must be 5 zeros.
If each of the 3 given zeros crosses the x- axis (which is what happens when it has an odd-numbered multiplicity), then there are 2 zeros missing.
The missing zeros are imaginary.
(3 zeroes × multiplicity of 1) + (2 imaginary)
3 + 2 = 5 [tex]\large\checkmark[/tex]
