Answer: C) 3 real zeros and 2 imaginary zeros
Explanation:
A function with a degree of 5 must have 5 total zeros.
It is given that there are 3 x-intercepts (zeros) and each has a multiplicity of 1, so there are 3 x 1 = 3 real zeros.
real + imaginary = 5
3 + imaginary = 5
imaginary = 2