Question

State how many imaginary and real zeros the function has. f(x) = x^5+7x^4+2x^3+14x^2+x+7

Answer 1

Since the degree of the equation is 5, then the number of zeros is also equal to 5. We can find for the 1st value of the root by trial and error. To do this, we find for the value of x which makes the equation equal to zero, that is:

x^5 + 7x^4 + 2x^3 + 14x^2 + x + 7 = 0          (find x)

By trial and error I made (which I will not show anymore), the possible root is -7. This is equivalent to x + 7.

To find more roots, we divide x^5 + 7x^4 + 2x^3 + 14x^2 + x + 7 by x + 7. The answer of the division results in x^4 + 2x^2 + 1 which is a perfect square trinomial in the form of:

(x^2 + 1)^2

Therefore the other 4 roots are:

x^2 + 1 = 0      ---> i, -i           

and

x^2 + 1 = 0      ---> i, -i

Therefore the answer is:

1 real and 4 imaginary zeros