Answer to old chestnut: (x-a)(x-b)...(x-z):

The answer is 0.

One of the terms is (x-x), which is zero, and zero times anything is zero, so the answer is zero.

(This is also an example of a namespace conflict, since in some circumstances you might really want to write something like this, but with the x in each term not equal to the x that gets subtracted. You could get around the conflict by replacing x with, say, α, but more often, this sort of thing is handled by using subscripted variables: (x-a0) times (x-a1) ... as far as you want to extend it.)

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