Find all real zeros of the function. I've been out of math classes for a long while so I'm pretty rusty.
Usually I can just look up a similar problem to what I'm working on and figure it out, but I'm not finding anything similar to this particular problem (just extremely simplified versions of it).
Find all real zeros of the function.
$$g(x)=4(x^2-4)(x+3)^2(x-3)$$
Can anyone explain this to me to help me solve it? I understand the basic idea of real zeros.
 A: For a product of some terms, only one of the terms has to be zero for the expression to be zero. This means that for $4(x^2-4)(x+3)^2(x-3)$, we can split this up further into $4(x+2)(x-2)(x+3)(x+3)(x-3) = 0$, which is also $(x+2)(x-2)(x+3)(x+3)(x-3) = 0$ dividing by $4$ on both sides.
Therefore, we can split the cases as follows:
$$\color{red} {(x+2)} = 0$$
$$\color{blue} {(x-2)} = 0$$
$$\color{green} {(x+3)} = 0$$  
$$\color{orange} {(x+3)} = 0$$  
$$(x-3) = 0$$
Can you find the roots from here?
A: The most important thing here is the Zero Factor Theorem, which is a fancy way of saying "if several things multiply to zero, then one of them is already zero."
Let's take a slightly different version of your problem (so that you can use yours for practice): say $h(x) = 7(x^2 - 9)(x + 2)^2(x - 5)$. If we have a zero - so $h(x) = 0$ - then it must be that one of those pieces is zero. So either $7 = 0$ (obviously not), $x^2 - 9 = 0$, $(x + 2)^2 = 0$, or $x - 5 = 0$. We'll deal with each possibility one at a time.
$7 = 0$: Obviously can't happen, so this doesn't give us any zeroes.
$x^2 - 9 = 0$: If $x^2 - 9 = 0$, then $x^2 = 9$, so $x = \pm 3$. That gives us the zeroes $-3$ and $3$.
$(x + 2)^2 = 0$: This can only happen if $x + 2 = 0$, so $x = -2$. That gives us the zero $-2$.
$x - 5 = 0$: If $x - 5 = 0$, then $x = 5$, giving us the zero $5$.
Thus the real zeroes of $h$ are $-3$, $3$, $-2$, and $5$.
