Question 1154198
 given zeros:

{{{x[1]=0}}} and 
{{{x[2]=2+5i}}}-> complex zeros always come in pairs, so you also have
{{{x[3]=2-5i}}}

{{{f(x)=(x-x[1])(x-x[2])(x-x[31])

{{{f(x)=(x-0)(x-(2+5i))(x-(2-5i))}}}

{{{f(x)=(x)(x-2-5i)(x-2+5i)}}}

{{{f(x)=(x)(x^2-2x+x*5i-2x+4-10i-x*5i+10i-25i^2)}}}

{{{f(x)=(x)(x^2-2x+cross(x*5i)-2x+4-cross(10i)-cross(x*5i)+cross(10i)-25(-1))}}}

{{{f(x)=(x)(x^2-4x+4+25)}}}

{{{f(x)=(x)(x^2-4x+29)}}}

{{{f(x)=x^3-4x^2+29x}}}


{{{ graph( 600, 600, -20, 20, -500, 500, x^3-4x^2+29x) }}}