Help with finding the real zeros of a polynomial $$P(x)=x^4-6x^3+4x^2+15x+4$$
Steps I took:
Possible zeros are:
$$(x+1)(x-1)(x+2)(x-2)(x+4)(x-4)$$
Used synthetic division to find which zero is an actual zero:
(I apologize for the graphical representation, I don't know how else to represent it on here)
$$-1\quad|\quad 1\quad -6\quad 4\quad 15\quad 4\\ ---------------\\ \quad \quad \quad \quad -1\quad 7-11-4\\ ---------------\\ \quad \quad 1\quad -7\quad 11\quad 4\quad 0$$
So the result is:
$$(x+1)(x^3-7x^2+11x+4)$$
Now I don't know how to factor this. Either I am not looking at this correctly or I made an amateur mistake somewhere up this point. 
Guide me in the right direction please. 

To check that $x=-1$ is a solution, all we need to do is to plug in, we don't need to divide by $x+1$. 
Note that $x=4$ also happens to be a solution. Now divide your polynomial by $(x+1)(x-4)$, that is, by $x^2-3x=4$.  You will get a quadratic as quotient, and you can use the Quadratic Formula on that.  
