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 
 A: Hint:$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
A: $x^3 + 64 = x^3+4^3 = (x+4)(x^2-4x+16) = (x+4)((x-2)^2+12) = 0 \to x+4 = 0 \to x= -4$. 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: $p(x) = x^3 + (\sqrt[3]{q})^3 = (x+\sqrt[3]{q})(\cdots)$
A: It's not half as complicated as the other answers make it look -- factoring the polynomial is (in this case) a pointless detour when you just want the roots.
You want to find $x$ such that $$x^3+64=0$$
Subtract 64 on both sides to get
$$ x^3 = -64 $$
We can stop here and say $x=\sqrt[3]{-64}$ ... but in case you don't know how to take cube roots of negative numbers, we can negate both sides:
$$ -(x^3) = 64 $$
and note that $-(x^3)$ happens to be the same as $(-x)^3$, so
$$ (-x)^3 = 64 $$
$$ -x = \sqrt[3]{64} = 4 $$
and finally negate once more to find
$$ x = -4 $$
A: 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.
