Question 1126930


 if {{{x[1]=6 +i}}} and {{{x[2]=7}}} are given zeros, there must be also zero {{{x[3]=6 -i}}} because complex zeros always coming in pairs

use zero product formula:

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

{{{f(x)=(x-(6 +i))(x-7)(x-(6 -i))}}}

{{{f(x)=(x-6 -i)(x-7)(x-6 +i)}}}

{{{f(x)=(x^2-6x -xi-7x+42+7i)(x-6 +i)}}}

{{{f(x)=(x^2-13x -xi+42+7i)(x-6 +i)}}}

{{{f(x)=(x^2-(13 +i)x+7(6+i))(x-6 +i)}}}

{{{f(x)=x^3 - 19 x^2 + 121x - 259}}}