Question 983477
the polynomial with leading coefficient {{{-2}}} that has degree {{{3}}} and roots {{{x[1]=2}}} and {{{x[2]=3i}}}

if the polynomial has {{{x[2]=3i}}}, then must have {{{x[3]=-3i}}} because complex roots always come in pairs

use zero product formula:

{{{f(x)=(x-x[1])(x-x[2])(x-x[3])}}}....since given that leading coefficient is {{{-2}}}, multiply product above by {{{-2)


{{{f(x)=-2((x-x[1])(x-x[2])(x-x[3]))}}}....substitute given roots

{{{f(x)=-2((x-2)(x-3i)(x-(-3i)))}}}

{{{f(x)=-2((x-2)(x-3i)(x+3i))}}}

{{{f(x)=-2((x-2)(x^2-(3i)^2))}}}

{{{f(x)=-2((x-2)(x^2-9(-1)))}}}

{{{f(x)=-2((x-2)(x^2+9))}}}

{{{f(x)=-2(x^3+9x-2x^2-18)}}}

{{{f(x)=-2x^3-18x+4x^2+36}}}

{{{f(x)=-2x^3+4x^2-18x+36}}}