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.)