How to determine $x$ and $y$ intercepts for $y = 4(x - 2)^2(x + 2)^3$ I need help to determine $x$ and $y$ intercepts for 
$$
y = 4(x - 2)^2(x + 2)^3
$$
I guess my first question is, do I need to get the equation into 
$$
ax^3 + bx^2 + cx + d 
$$
form before starting?
 A: The $x$-intercepts are defined to be the points where $y=0$.  Since what you have is factored, it is easy to see where $y=0$.  If $y=0$ then one of the factors must be $0$.  The $y$-intercept is where $x=0$.  If it exists, it is unique.  Just plug in $x=0$ and see what you get.
A: That would be counterproductive.
The $x$-intercepts are solutions to $y=0$, and your equation is set up so that this is easy to solve.
For the $y$-intercept, just set $x=0$ and calculate $y$.
A: Multiplying is easy; factoring is hard.  (Believe it or not, that's why it's hard for identity thieves to steal your money in properly secured e-commerce transactions.  Unless maybe the thief is the Prince of Mongolia and you're inexperienced.)
If you had a polynomial in the form $ax^3 + bx^2 + cx +d$, you'd have to factor it to find the $x$-intercepts.  This one you've already got in factored form.
A: The x-intercepts can be calculated like this:
$$ y = 4(x - 2)^2(x + 2)^3 $$
$$ y = 0$$
$$ a \cdot b \cdot c = 0 \rightarrow a = 0 \vee b=0 \vee c=0 $$
$$ 4 = 0 \vee (x-2)^2 = 0 \vee (x+2)^3 = 0$$
$$ None \vee x = 2 \vee x \ -2$$
The y-intercept can be found by substituting $0$ for $x$ in $4(x - 2)^2(x + 2)^3$.
A: The zeroes of the equation $y = 4(x-2)^2(x+2)^3$ are clearly defined as $x = +2, -2$ which everyone stated. Another idea which I think is interesting is the equations behaviour around these roots. The factor $(x-2)^2$ has an order of $2$ therefore near this root the equation will "bounce" or act quadratically. The factor $(x+2)^3$ has an order of 3 therefore the equation will act cubically near this point, with an inflection at the root.
