Quadratic equation / why does $x(x-2)=0$ imply $x = 0 \lor x = 2$? I feel silly asking such elementary questions, but hopefully this is appropriate for math.stackexchange.
I'm studying to take calculus next semester but I haven't done any math in a long time, so I've been trying to brush up on my algebra and my trig. I'm stuck on this problem here: 
$$
(4 x-2)^2-2 (4 x-2) = 15
$$
I was able to get to this step:
$$
8-24 x+16 x^2 = 15
$$
But after I'm confused on what to do after this step.
Also(a related question): I got some help with this equation earlier:
$$(u+1)/(u+4)+1 = (u-5)/(u-4)$$
The last step ended up being $$x (x -2) = 0$$
The explanation I recieved was that "this implied that $x = 0$ or $x = 2$" I was wondering if someone can help me understand why this is so, I feel as if I'm missing a vital piece of information  preventing me from understanding these problems.
I'd be much obliged if someone can point me in the right direction.
Here are the wolfram links for both equations:
http://www.wolframalpha.com/input/?i=%284x-2%29%5E2-2%284x-2%29%3D15
http://www.wolframalpha.com/input/?i=%28u%2B1%29%2F%28u%2B4%29%2B1+%3D+%28u-5%29%2F%28u-4%29
 A: It's one of the basic features of numbers that if you multiply two numbers and the result is $0$, then at least one of the factors must have been $0$. Namely if $ab=0$ and $a$ is not $0$, then we're allowed to divide both sides of the equation by $a$ and get $b=\frac0a=0$. So either $a$ is $0$ or $b$ is.
Therefore $x(x-2)=0$ implies that $x=0$ or $x-2=0$ and the latter is just a different way to write $x=2$.
A: We consider only the problem of finding the solutions of $(4 x-2)^2-2 (4 x-2) = 15$. You multiplied out correctly.  We will look at the equation you got a little later.  But first, we do the problem in a more lazy (and efficient) way.  
It seems natural to let $u=4x-2$.  Then our equation can be rewritten as
$$u^2-2u-15=0.$$
Possibly we can find the solutions by inspection.  Or else we can factor the quadratic, obtaining
$$u^2-2u-15=(u-5)(u+3).$$
Note that $(u-5)(u+3)=0$ when $u=5$, when $u=-3$, and for no other value of $u$.  It follows that $x$ is a solution of the original equation when $4x-2=-3$ and when $4x-2=5$, and for no other value of $x$.  
We now have a couple of linear equations for the roots. To solve $4x-2=-3$, rewrite as $4x=(-3)+2=-1$, and divide both sides by $4$ to get $x=-\dfrac{1}{4}$. Similarly, the equation $4x-2=5$ has the solution $x=\dfrac{7}{4}$.
Instead of factoring $u^2-2u-15$, we could use the Quadratic Formula. For details, please see below.
Another way: Expand, as you did, and bring all terms to the left-hand side. We obtain the equation
$$16 x^2-24x-7 =0.$$
It turns out that the quadratic polynomial on the left can be factored without too much trouble.  (Many quadratic polynomials in school exercises can be factored in a simple way. One cannot count on that in the real world!)  So we recall the standard formula for solving the quadratic equation $ax^2+bx+c=0$ (where $a \ne 0$).
The solutions are 
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
For a derivation of this formula, any source will do. I even have one on Stack Exchange. The formula is so frequently useful that most students remember it. 
Now calculate, with $a=16$, $b=-24$, and $c=-7$. It turns out that $b^2-4ac=1024=(32)^2$, so the roots are not complicated.
Comment: Let $P(x)$ be a polynomial with integer coefficients. Suppose that the  highest degree term is $ax^n$, and the constant term is $c$.  Then any rational root of $P(x)=0$ must be of the form $p/q$, where $p$ is a (possibly negative) factor of $c$, and $q$ is a positive factor of $a$.  (This result is often called the Rational Roots Theorem.)  
So if we are looking for the rational roots of $16x^2-24x-7=0$, the only possible candidates are $\pm\frac{1}{1}$, $\pm\frac{1}{2}$, $\pm\frac{1}{4}$, $\pm\frac{1}{8}$, $\pm\frac{1}{16}$, $\pm\frac{7}{1}$, $\pm\frac{7}{2}$, $\pm\frac{7}{4}$, $\pm\frac{7}{8}$,  and $\pm\frac{7}{16}$.  We can just try all these, to see which, if any, work.  It isn't quite as unpleasant as it looks. However, it is not really a suitable approach to quadratic equations.  After all, the roots might not be rational.  And the Quadratic Formula is simple to use. 
