Question 357714
Since {{{2i}}} is a root, then {{{-2i}}} is also a root.
{{{f(x)=a(x-2)(x+2i)(x-2i)}}}
{{{f(x)=a(x-2)(x^2+4)}}}
{{{f(1)=a(1-2)(1+4)=10}}}
{{{a(-1)(5)=10}}}
{{{a=-2}}}
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{{{f(x)=-2(x-2)(x^2+4)}}}
{{{f(x)=-2(x^3+4x-2x^2-8)}}}
{{{highlight(f(x)=-2x^3+4x^2-8x+16)}}}