Question 1027930
The assumption is that the coefficients of the cubic polynomial are real so the complex conjugate is also a root.
{{{f(x)=a(x-3)(x-4i)(x+4i)}}}
{{{f(x)=a(x-3)(x^2+16)}}}
Now use the point to solve for {{{a}}}.
{{{f(1)=a(1-3)(1^2-16)}}}
{{{68=a(-2)(-15)}}}
{{{a=68/30}}}
{{{a=34/15}}}
.
.
{{{f(x)=(34/15)(x-3)(x^2+16)}}}