Answer
$f(x)=x^3-x^2+25x-25$
Work Step by Step
If $5i$ is a zero of $f$ then the complex conjugate $-5i$ is also a zero.
$f(x)=a(x-1)(x-5i)[x-(-5i)]$
$f(x)=a(x-1)(x-5i)(x+5i)$
$f(x)=a(x-1)[x^2-(5i)^2]$
$f(x)=a(x-1)(x^2+25)$
$f(x)=a(x^3+25x-x^2-25)$
We can choose any value for $a$. So, let's make $a=1$
$f(x)=x^3-x^2+25x-25$
