Respuesta :
x = 4; x = 2i; x = -2i
[tex]f(x) = (x - 4)(x - 2i)[x - (-2i)] = (x - 4)\underbrace{(x-2i)(x+2i)}_{a^2-b^2=(a-b)(a+b)}\\\\=(x-4)[x^2-(2i)^2]=(x-4)(x^2-2^2i^2)=(x-4)[x^2-4(-1)]\\\\=(x-4)(x^2+4)=(x)(x^2)+(x)(4)-(4)(x^2)-(4)(4)\\\\=x^3+4x-4x^2-16=\boxed{x^3-4x^2+4x-16}[/tex]
[tex]f(x) = (x - 4)(x - 2i)[x - (-2i)] = (x - 4)\underbrace{(x-2i)(x+2i)}_{a^2-b^2=(a-b)(a+b)}\\\\=(x-4)[x^2-(2i)^2]=(x-4)(x^2-2^2i^2)=(x-4)[x^2-4(-1)]\\\\=(x-4)(x^2+4)=(x)(x^2)+(x)(4)-(4)(x^2)-(4)(4)\\\\=x^3+4x-4x^2-16=\boxed{x^3-4x^2+4x-16}[/tex]