## Algebra 2 (1st Edition)

$f(x)=x^3-9x^2+28x-30$
The coefficients are rational, thus by the complex conjugates theorem, $3+i$ is also a factor. Using the Factor Theorem if $x=a$ is a factor, then $(x-a)$ is a factor of $f(x)$. Hence here $f(x)=(x-3)(x-(3+i))(x-(3-i))=(x-3)(x-3-i)(x-3+i))=(x-3)((x-3)^2-i^2)=(x-3)(x^2-6x+10)=x^3-6x^2+10x-3x^2+18x-30=x^3-9x^2+28x-30$