Question 1178286

given:

Zeros : {{{1}}} , {{{2+i }}}, {{{2-i}}}


using zero product rule we have

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

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

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

{{{f(x)=(x-1)(x^2 - 4x + 5)}}}

{{{f(x)=x^3 - 5x^2 + 9x - 5}}}