Answer
$ x=4, -2 \pm \sqrt 5$
Work Step by Step
When a polynomial $ f(x)$ is divided by $(x-a)$, then the remainder is $ r=f(a)$
We are given that $ x=4$ is a root of $ x^3-17x+4=0$ and this means $(x-4)$ is a factor of $ x^3-17x+4=0$.
So, $ x^3-17x+4=(x-4)\times x^2+4x^2-17x+4$
or, $=(x-4) (x^2+4x)-x+4$
or, $=(x-4) [(x+2)^2-5]$
We need to solve for $ x $, so set each factor of $[(x+2)^2-5]$ and $ x-4$ equal to $0$.
so, $[(x+2)^2-5] \implies x= -2 \pm \sqrt 5$
Thus, the roots of $ x^3-17x+4=0$ are $ x=4, -2 \pm \sqrt 5$
