## Algebra and Trigonometry 10th Edition

$f(x)=x^3-4x^2+9x-36$
If $-3i$ is a zero of $f$ then the complex conjugate $3i$ is also a zero. $f(x)=a(x-4)(x-3i)[x-(-3i)]$ $f(x)=a(x-4)(x-3i)(x+3i)$ $f(x)=a(x-4)[x^2-(3i)^2]$ $f(x)=a(x-4)(x^2+9)$ $f(x)=a(x^3+9x-4x^2-36)$ We can choose any value for $a$. So, let's make $a=1$ $f(x)=x^3-4x^2+9x-36$