Question 1004969
{{{x^3+3x^2+16x-20=0}}}

using zero product theorem, we can write the 3rd degree function as 

{{{(x-x[1])(x-x[2])(x-x[3])=0}}}

given:
{{{x[1]=-2+4i}}}...if you have this root, you have to have {{{x[2]=-2-4i}}} too (complex roots always in pairs)

then we have


{{{(x-(-2+4i))(x-(-2-4i))(x-x[3])=0}}}

{{{(x+2-4i)(x+2+4i)(x-x[3])=0}}}

{{{(x^2+2x+4xi+2x+4+8i-4xi-8i-16i^2)(x-x[3])=0}}}

{{{(x^2+2x+cross(4xi)+2x+4+cross(8i)-cross(4xi)-cross(8i)-16(-1))(x-x[3])=0}}}

{{{(x^2+2x+2x+4+16)(x-x[3])=0}}}

{{{(x^2+4x+20)(x-x[3])=0}}}

so, 

{{{x^3+3x^2+16x-20=(x^2+4x+20)(x-x[3])}}}

{{{(x^3+3x^2+16x-20)/(x^2+4x+20)=(x-x[3])}}}

-----------------({{{x-1}}}
{{{(x^2+4x+20)}}}|({{{x^3+3x^2+16x-20}}})
-------------------{{{x^3+4x^2+20x}}}........subtract
--------------------{{{0-x^2-4x}}}...........bring down {{{-20}}}
-------------------...{{{-x^2-4x-20}}}
-------------------...{{{-x^2-4x-20}}}...subtract
-------------------...........{{{0}}}......reminder

so, third product is {{{(x-x[3])=x-1}}}=>{{{x[3]=1}}}

and your equation is factored as following:

{{{(x+2-4i)(x+2+4i)(x-1)=0}}}


and, roots are:
{{{x[1]=-2+4i}}},
{{{x[2]=-2-4i}}}, and 
{{{x[3]=1}}}