Answer :
[tex]f(x)=(x-1)(x-1)(x+2i)(x-2i)
\\f(x)=(x^2-2x+1)(x^2-(2i)^2)
\\f(x)=(x^2-2x+1)(x^2-4i^2)
\\f(x)=(x^2-2x+1)(x^2-4(-1))
\\f(x)=f(x)=(x^2-2x+1)(x^2+4)
\\f(x)=x^4-2x^3+x^2+4x^2-8x+4
\\f(x)=x^4-2x^3+5x^2-8x+4[/tex]
zeros of the function
[tex]f(x)=(x-1)(x-1)(x+2i)(x-2i) \\x-1=0 \Rightarrow x=1 \\x+2i=0 \Rightarrow x=-2i \\x-2i=0 \Rightarrow x=2i[/tex]
zeros of the function
[tex]f(x)=(x-1)(x-1)(x+2i)(x-2i) \\x-1=0 \Rightarrow x=1 \\x+2i=0 \Rightarrow x=-2i \\x-2i=0 \Rightarrow x=2i[/tex]