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. 

I think I understand your graphical representation. So you have found that $-1$ is a root, which is correct. It is also nice that you have divided out $(x+1)$. Now you have either one or three real roots left to find. Here is how I would recommend proceeding. You have a list of six possible things that could be zeroes. One of them, which I imagine you would inevitably try is $x=4$. This means you can factor out $(x+1)(x-4)$ from your polynomial, and rewrite $$P(x)=(x+1)(x-4)(x^2-3x-1)$$ Now all you have to do is use the quadratic formula to solve $$x^2-3x-1=0$$ for your last two roots and you are done.
