Question 966814
{{{(x-4)(x-(-10))(x-(3+4i))(x-(3-4i))}}}


{{{(x-4)(x+10)(x-3-4i)(x-3+4i)}}}


{{{(x-4)(x+10)((x-3)-4i)((x-3)+4i)}}}


{{{(x-4)(x+10)((x-3)^2-(4i)^2)}}}


{{{(x-4)(x+10)(x^2-6x+9-16i^2)}}}, and remember, {{{i^2=-1}}};


{{{(x-4)(x+10)(x^2-6x+9+16)}}}


{{{(x-4)(x+10)(x^2-6x+25)}}}


Still not yet in general form, but you see the main process.  The three given zeros, one of them being complex, you need also the conjugate of that complex zero; so FOUR zeros for the polynomial.  Setup the factors, and the rest is algebraic steps.