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

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.

1 ?

step-by-step explanation:

u said 1+1+1+1+1+1+1+1+1+1+1+1*0 so ur adding 1 but then multiplying by 0 so its 0 but then u said 1*0+1 so now ur adding 1 so the answer is juan ?

~leah

10 - 8 = -2

  8 - 6 = -2

etc, so -2 is the common difference

step-by-step explanation: