Given that zeros of a function are the solutions of fx0 and

Given that zeros of a function are the solutions of f(x)=0, and the zeros can be real or imaginary,determine the five zeros of the 5th degree polynomial: f(x)= (x-2)^2 (x+5)(x^2+7)

Solution

f(x)= (x-2)^2 (x+5)(x^2+7)

Given that zeros of a function are the solutions of f(x)=0

(x-2)^2 (x+5)(x^2+7) =0

(x -2) =0 ---> x = 2 ( repeated root , multiplicity 2)

(x +5) = 0 ----> x = -5

(x^2 +7) =0 ----> x = + /- i*sqrt(7)

So, five zeros are x = 2 , 2 , -5 , i*sqrt7 , -i*sqrt7