What are zero divisors used for?

Trying to shed additional light by means of an example involving a most typical deduction where the presence/absence of zero divisors plays a role. Let's imagine that we need to solve the equation $$x^2-x=0.$$ Before we can make any progress on this, we need to know a bit more. What kind of an object is $x$? Is it a number (real/complex/rational)? Is it a matrix? Is it an element of some other universe, where the equation makes sense? A couple things are certain. Whatever $x$ is, we need the ability to multiply it with itself (otherwise $x^2$ does not exist) as well as subtract it from its square. Furthermore, this universe must have an object called zero. Let's assume that the universe is at least a ring, because then we have all of the above operations available. In a ring we have a unit (a neutral element for multiplication, "$1$") and the distributive law. This allows us to rewrite the equation as $$ 0=x^2-x=x^2-1\cdot x=(x-1)x. $$ Well, does that help? Depends! If $x$ is an element of any of the school level number systems, we know that the system has no zero divisor. Meaning that if $ab=0$ in such a system, we are allowed to deduce that either $a=0$ or $b=0$. If our $x$ comes from such a system, we can make swift progress with our equation and conclude that either $$ x-1=0\qquad\text{or}\qquad x=0. $$ This tells us that either $x=1$ or $x=0$ (standard applications of ring axioms).

So we have completely solved the equation PROVIDED that $x$ resides in a ring that does not have zero divisors. This is just an abstraction of the techniques we learned in school, but there no attention to the zero divisor concept is usually given, even though it did play a key role. That is all fine, because drawing attention to the zero divisor concept begs the question: "Are there meaningful systems with zero divisors?"

Let us next assume that $x$ is an element of the residue class ring $\mathbb{Z}_6$. Students might nevertheless want to use the above technique. But some wiseguy will notice that $x=\overline{3}$ is also a solution, as $x^2=\overline{9}=\overline{3}$. What was wrong with the earlier approach? Let's check: $$ (x-1)x=\overline{(3-1)}\cdot\overline{3}=\overline{2}\cdot\overline{3}=\overline{6}=0. $$ Lo and behold, in this ring a prodct can be zero without either factor being so. But the need to solve our equation still is there. If we are not allowed to use the familiar trick, then what can we do? Quadratic formula? That requires an ability to calculate square roots, and an ability to divide by two. Without going into details I will just state that both of these are suspect in more general rings. Also the absence of zero divisors is hidden in the derivation of the quadratic formula. So what? Thankfully this ring is finite, so we can get away by simply trying all the possible values of $x$. Such checking reveals that the residue class $x=\overline{4}$ is also a solution.

But it gets worse! If $x$ is a $2\times2$ matrix with real entries, then in addition to the "obvious" solutions $$ x=\left(\begin{array}{rr}1&0\\0&1\end{array}\right),\qquad\text{and}\qquad x=\left(\begin{array}{rr}0&0\\0&0\end{array}\right) $$ we also have solutions like $$ x=\left(\begin{array}{rr}1&0\\0&0\end{array}\right),\qquad\text{and}\qquad x=\left(\begin{array}{rr}0&0\\0&1\end{array}\right).$$ And that's not all. I invite you to check that the matrix $$ x=\left(\begin{array}{cc}\cos^2\alpha&-\cos\alpha\sin\alpha\\-\cos\alpha\sin\alpha&\sin^2\alpha\end{array}\right) $$ is a solution of the equation $x^2-x=0$ for all values of the angle $\alpha$.

So the main "motivation" to study the concept of zero divisors is, as André pointed out, to know when they are absent. A simple task may become quite harrowing, if we don't know about the possibility of zero divisors.

Above we saw that a quadratic equation may have four or infinitely many solutions, if the ring has zero divisors. With a little bit of extra work we can prove the familiar result that a degree $n$ equation in a commutative ring without zero divisors has at most $n$ solutions. The reason for adding the commutativity assumption is a bit subtle, and I won't get into that. If I managed to awake your curiosity about that, I advice you to take a peek at this question for an example of a quadratic equation with infinitely many solutions in a non-commutative ring without zero divisors. I also recommend the answer by Arturo Magidin, but it does require some familiarity with ring concepts.

Just to add something new to what is already said in the previous answers: in some cases you have "useful" zero-divisors. For example, let $V$ be a vector space and let $W$ be a subspace. Consider $A=\mathrm{End}(V)$, the ring of linear endomorphisms of $V$, endowed with the usual pointwise sum and the composition product. You can define a nontrivial endomorphism $\pi : V \to V$ in $A$ such that $\pi(V) = W$ and $\pi_{|W} = 1_W$. This is what you call a "projection", and it is idempotent, that is $\pi \circ \pi = \pi$. It is also a zerodivisor: if it weren't, then you would have $\pi \circ(1-\pi)=0$ and then $\pi = 0$ or $\pi = 1_V$. It should be pointed out that the existence of nontrivial idempotents on a ring is useful to obtain decomposition results for modules over the given ring. Read, for example, this article:

http://en.wikipedia.org/wiki/Idempotent_element#Role_in_decompositions