Find the zeros of $f(x)=x^3+64$ $$f(x)=x^3+64$$ 
Again, I am really not sure how to do this I tried to factor but it clearly was not the right answer 

$x^3 + 64 = x^3+4^3 = (x+4)(x^2-4x+16) = (x+4)((x-2)^2+12) = 0 \to x+4 = 0 \to x= -4$. I suppose you want real roots. If you want complex roots, then $(x-2)^2 = -12 = (2\sqrt{3}\cdot i)^2 \to x = 2 \pm 2\sqrt{3}\cdot i$.
In general, if you are interested in finding the roots of $p(x) = x^3 + q$, then you first write: $p(x) = x^3 + (\sqrt[3]{q})^3 = (x+\sqrt[3]{q})(\cdots)$
