Question 781983


{{{ax^3+bx^2+cx+d=0}}} where {{{a=1}}}

so, your polynomial will be {{{x^3+bx^2+cx+d=0}}}

use zero product rule since you are given zeros {{{x[1]=2}}},{{{x[2]=-5i}}} and {{{x[3]=5i}}}

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

{{{(x-2)(x-(-5i))(x-5i)=0}}}

{{{(x-2)(x+5i)(x-5i)=0}}} ......{{{(x+5i)(x-5i)=x^2-(5i)^2=x^2-(25(-1))=x^2+25}}}

{{{(x-2)(x^2+25)=0}}}

{{{x^3+25x-2x^2-50=0}}}

{{{x^3-2x^2+25x-50=0}}}