Question 914543
 {{{1+i}}} =>complex root 
and 
{{{-5}}} real root

Complex roots always come in complex conjugate pairs; so we have {{{1+i}}} and {{{1-i}}}

{{{f(x)=(x-(1+i))(x-(1-i))(x-(-5))}}}


{{{f(x)= (x-1-i)(x-1+i) (x+5)}}} ....multiply


{{{f(x)=(x^2-x+cross(x*i)-x+1-cross(i)-cross(xi)+cross(i)-i^2)(x+5)}}}


{{{f(x)=(x^2-2x+1-(-1))(x+5)}}}


{{{f(x)=(x^2-2x+1+1)(x+5)}}}


{{{f(x)= (x^2-2x+2)(x+5) }}}


{{{f(x)=x^3+5x^2 -2x^2-10x +2x+10}}}

{{{f(x)=x^3+3x^2-8x+10}}}


{{{ graph( 600, 600, -10, 10, -10, 15, x^3+3x^2-8x+10) }}}