Question 1147927
{{{ 2 + I }}} is also a root
{{{ f(x) = ( x - ( 2 - i ) )*( x - ( 2 + i ) )*( x - 3 )*( x + 1 ) }}}
{{{ f(x) =( x^2 - ( 2 - i + 2 + i )*x + 5 )*( x - 3 )*( x + 1 ) }}}
{{{ f(x) = ( x^2 - 4x + 5 )*( x^2 - 2x - 3 ) }}}
{{{ f(x) = x^4 - 4x^3 + 5x^2 - 2x^3 + 8x^2 - 10x - 3x^2 + 12x - 15 }}}
{{{ f(x) = x^4 - 6x^3 + 10x^2 + 2x - 15 }}}
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get a 2nd opinion if needed