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 

At the precalculus level, any cubic polynomial (i.e. a function of the form $f(x) = ax^3 + b x^2 + cx + d$) that you are asked to find the roots of will almost certainly have one "obvious root".
In this example, can you find a value of $x$ such that $f(x) = 0$ - i.e. $$x^3 = -64$$
Once you have this one root, let's call it $a$, you can use this root to find the other roots by factorising the cubic. Since $f(a) = 0$, by the Remainder Theorem, we know that $f$ has $(x-a)$ as a factor. So find $b, c$ such that $$f(x) = (x-a)(x^2 +bx + c)$$ and you're left with a quadratic to factorise. The roots of $f$ are then just $a$ and the roots of the quadratic equation $x^2 + bx +c$, which you can solve in the usual way.
