Question

(×2+1)(x3+2x)(x2+4x-16i-4xi) Solve for the roots in equation

Answer 1

$\bf (x^2+1)(x^3+2x)(x^2+4x-16i-4xi)=0\\\\ -------------------------------\\\\ x^2+1=0\implies x^2=-1\implies x=\sqrt{-1}\implies x=i\\\\ -------------------------------\\\\ x^3+2x=0\implies x(x^2+2)=0\implies \begin{cases} x=0\\ -------\\ x^2+2=0\\ x^2=-2\\ x=\sqrt{-2}\\ x=\sqrt{2\cdot -1}\\ x=\sqrt{2}\cdot \sqrt{-1}\\ x=i~\sqrt{2} \end{cases}\\\\ -------------------------------$

$\bf x^2+4x-16i-4xi=0\implies (x^2+4x)~-~(16i+4xi)=0 \\\\\\ (x^2+4x)~-~(4xi+16i)=0\implies x(x+4)~-~4i(x+4)=0 \\\\\\ \stackrel{common~factor}{(x+4)}(x-4i)=0\implies \begin{cases} x+4=0\\ x=-4\\ -----\\ x-4i=0\\ x=4i \end{cases}$