Question 1078910

given:

 {{{1}}} & {{{1+2i}}}

if we have {{{1+2i}}}, than we have also {{{1-2i}}}, complex solutions come always in pairs

so, we use zero product formula to find equation:

{{{f(x)=(x-x[1])(x-x[2])(x-x[3])}}}....plug in {{{x[1]=1}}}, {{{x[2]=1+2i}}}, and {{{x[3]=1-2i}}}


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

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

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

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

{{{f(x)=x^3 - 3x^2 + 7x - 5}}}


{{{ graph( 600, 600, -10, 10, -10, 10, x^3 - 3x^2 + 7x - 5) }}}