Question 317254
Since {{{i}}} is a root, then {{{-i}}} is also a root since complex roots only occur in complex conjugate pairs.
.
.
.
{{{f(x)=A(x-i)(x+i)(x-3)}}}
{{{f(x)=A(x^2+1)(x-3)}}}
.
.
Use the point to solve for A.
.
.
{{{f(2)=A(4+1)(2-3)=10}}}
{{{A(5)(-1)=10}}}
{{{A=-2}}}
.
.
.
{{{f(x)=-2(x^2+1)(x-3)}}}
{{{f(x)=-2(x^3-3x^2+x-3)}}}