State how many imaginary and real zeros the function has.
f(x) = x5 + 7x4 + 2x3 + 14x2 + x + 7
a. 3 imaginary; 2 real
b. 4 imaginary; 1 real
c. 0 imaginary; 5 real
d. 2 imaginary; 3 real

Hello,
Degre 5==> even number of imaginary roots and odd number of reals.

Real roots must be a divisor of 7.
Using Excel, we find

x  P(x) 
 -7 0
-6 1369
-5 1352
-4 867
-3 400
-2 125
-1 24
 0 7
 1 32
 2 225
 3 1000
 4 3179
 5 8112
 6 17797
 7 35000

So: one real root: 7 and 4 imaginary roots.

Answer B