Question 1137185


{{{n=3}}}
{{{3}}} and {{{4i}}} are zeros->{{{-4i}}} is also zero (complex zeros always come in pairs)

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

{{{f(x)=a(x-3)(x-4i)(x+4i)}}}

{{{f(x)=a(x-3)(x^2-(4i)^2)}}}

{{{f(x)=a(x-3)(x^2-16i^2)}}}

{{{f(x)=a(x-3)(x^2-16(-1))}}}

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

{{{f(x)=a(x^3+16x-3x^2-48)}}}

{{{f(x)=a(x^3-3x^2+16x-48)}}}

use
{{{f(1)= -102}}} to find {{{a}}}

{{{-102=a(1^3-3*1^2+16*1-48)}}}

{{{-102=-34a}}}

{{{a=-102/-34}}}

{{{a=3}}}


{{{f(x)=3(x^3-3x^2+16x-48)}}}

{{{f(x)=3x^3-9x^2+48x-144)}}}-> your solution