Finding rational of zeros and then writing in factored form For an assignment in my Algebra 2 class we have to find the rational zeros of the equation then write it in factored form.
I am a little lost and don't really get it
Equation: 

$h(x)=x^3-5x^2+2x+8$

I have tried first getting the rational zeros and got 

$1, 2, 4, 8$

Then using synthetic substition

$2$ | $1$ $-5$ $2$ $8$
____2 -6 -8____

From that i got

$1x^3$ $-3$ $-4$ and finally $0$

Then wrote it into a function

$h(x)=(x-2)(x^2-3x-4x)$

Which i don't know if it is correct, i also don't know what they mean by factor it
It would be helpful if you could show all you steps so i can actaully learn how to do it myself.
Thanks in advance!

You're doing everything right so far! Just factor the quadratic $x^2-3x-4$ to completely factor $x^3-5x^2+2x+8$.

Although, you could do it this way:
$$\begin{align*} & x^3-5x^2+2x+8\\ & =x^3-4x^2-x^2+2x+8\\ & =x^2(x-4)-(x-4)(x+2)\\ & =(x-4)(x^2-x-2)\\ & =(x-4)(x-2)(x+1)\end{align*}\tag{1}$$
But not many people might've seen the first step...
