Question 1161653
 given :

{{{n=3}}}->degree

zeros:
{{{x[1]=4}}} 
{{{x[2]=4i}}} -> complex solutions always come in pairs, so you also have
{{{x[3]=-4i}}} 


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


{{{f(x)=a(x-4)(x-4i)(x-(-4i))}}}


{{{f(x)=a(x-4)(x-4i)(x+4i)}}}->{{{(x-4i)(x+4i)=x^2-(4i)^2=x^2-16(-1)=x^2+16}}}


{{{f(x)=a(x-4)(x^2+16)}}}


{{{f(x)=a(x^3-4x^2+16x-64)}}}


then, use given


{{{f(1)= -102}}} to calculate {{{a}}}

{{{ -102=a(1^3-4*1^2+16*1-64)}}}

{{{ -102=-51a}}}

{{{ a=-102/-51}}}

{{{ a=2}}}

so, your equation is:

{{{f(x)=2(x^3-4x^2+16x-64)}}}

{{{f(x)=2x^3-8x^2+32x-128}}}