Stating that either root is zero in solving a quadratic equation Let's say we have a simple quadratic equation $x^2 - 3x = 0$. To solve, we will factor $x$ out i.e. $x(x-3)=0$, after which we will state $x = 0$ or $(x-3) = 0$. My question is, why is there no third "option" where we say "or both". Isn't it possible for both "portions" (i.e. $x$ and $(x-3)$) to both be equal to zero? After all, if most quadratic equations would have two roots, then both $x = 0$ and $x = 3$ are the roots thus both $x = 0$ and $(x-3) = 0$ are true! 
[This question on the use of the word "or" applies to polynomial equations with degrees 3 and above too of course, but I'm choosing a quadratic one as it is the simplest case possible.]

If you write a polynomial $p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$, then by a theorem (the Fundamental Theorem of Algebra), you can completely factor the polynomial as $p(x) = a_n(x - b_1)(x-b_2)\cdots(x - b_n)$ where the $b_j$'s are the (potentially complex) roots.
Then $p(x) = 0$ precisely when $x = b_j$ for at least one $j, 1 \leq j \leq n$.  With this phrasing, the "at least one" covers the issue you're talking about.  Note, however, if $x = b_j$ and $x = b_k$ that it follows that $b_j = b_k$, so two or more of the linear facts $(x - b_j), (x - b_k)$ will be zero simultaneously if and only if $b_j = b_k$.
So in your example, it can never be the case that for some $x$, both $x - 0 = 0$ and $x - 3 = 0$.  But if you take the polynomial $x^2 - 2x + 1 = (x-1)(x-1)$, then for $x = 1$ you have that both are simultaneously $0$.
