Chapter 5 Polynomials and Polynomial Functions - 5.7 Apply the Fundamental Theorem of Algebra - Guided Practice for Example 3 - Page 381: 6

$f(x)=x^3-6x^2+4x+16$

The coefficients are rational, thus by the irrational conjugates theorem, $1-\sqrt5$ 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 $f(x)=(x-4)(x-(1+\sqrt5))(x-(1-\sqrt5))=(x-4)(x-1-\sqrt5))(x-1-\sqrt5))=(x-4)((x-1)^2-5)=(x-4)(x^2-2x-4)=x^3-2x^2-4x-4x^2+8x+16=x^3-6x^2+4x+16$