Answer:
B. 4 imaginary; 1 real
Step-by-step explanation:
Given the polynomial:
x^5 + 7*x^4 + 2*x^3 + 14*x^2 + x + 7
it can be reordered as follows
(x^5 + 2*x^3 + x ) + (7*x^4 + 14*x^2 + 7)
Taking greatest common factor at each parenthesis
x*(x^4 + 2*x^2 + 1) + 7*(x^4 + 2*x^2 + 1)
Taking again the greatest common factor
(x + 7)*(x^4 + 2*x^2 + 1)
Replacing x^2 = y in the second parenthesis
(x + 7)*(y^2 + 2*y + 1)
(x + 7)*(y + 1)^2
Coming back to x variable
(x + 7)*(x^2 + 1)^2
There are two options to find the roots
(x + 7) = 0
or
(x^2 + 1)^2 = 0 which is the same that (x^2 + 1) = 0
In the former case, x = -7 is the real root. In the latter, (x^2 + 1) = 0 has no real solution. Therefore, there is only 1 real root in the polynomial.
