Question 963221

the cubic equation that has {{{x[1]=-1}}} and {{{x[2]=2i}}} as roots, have also {{{x[3]=-2i}}} because complex root come always in pairs

now, since we have all roots, we can use zero product rule to see which cubic equation  is right answer

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

{{{y=(x-(-1))(x-2i)(x-(-2i))}}}

{{{y=(x+1)(x-2i)(x+2i)}}}

{{{y=(x+1)(x^2-(2i)^2)}}}

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

{{{y=(x+1)(x^2-(4(-1)))}}}

{{{y=(x+1)(x^2-(-4))}}}

{{{y=(x+1)(x^2+4)}}}

{{{y=x^3+4x+x^2+4}}}

{{{y=x^3+x^2+4x+4}}}....=> so, answer is {{{D}}}

or, you can do it this way:

{{{y=x^3+ x^2 + 4x + 4}}}....to find zeros, set {{{y=0}}} and factor completely

{{{0=(x^3+ 4x)+ (x^2  + 4)}}}

{{{0=x(x^2+ 4)+ (x^2  + 4)}}}

{{{(x+ 1) (x^2  + 4)=0}}}

solutions:

if {{{(x+ 1)}}} => {{{x=-1}}}
if {{{(x^2  + 4)=0}}} => {{{x^2 =- 4}}}=>{{{x=sqrt(-4)}}}=>{{{x=2i}}} or {{{x=-2i}}}

=> one more time we got that answer is {{{D}}}