Answer
$\{-1, 1, 4\}$
Work Step by Step
Step 1. Given the equation $x^3-4x^2-x+4=0$, factor by grouping, we have $x^2(x-4)-(x-4)=0$, $(x-4)(x^2-1)=0$, and $(x-4)(x+1)(x-1)=0$
Step 2. Solve the above equation to get $x=\pm1, 4$ or in set form as $\{-1, 1, 4\}$
