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Zeros of a PolynomialThe following demonstrates a couple of simple, yet extremely important properties of polynomials. In the section on ring homomorphisms we saw the following example: Let R be any commutative ring, X indeterminate over R,
f(a) = a_{0} + a_{1}a + ... + a_{n}a^{n}.
Let
_{a}(f ) = f(a) (f R[X]).
Then _{a} is a homomorphism. Indeed, if
g = b_{0} + b_{1}X + ... + b_{n}X^{n} + ... R[X]
then
_{a}(f + g) = _{a}((a_{0}+b_{0})+(a_{1}+b_{1})X+...+(a_{n}+b_{n})X^{n}+...)
= (a_{0}+b_{0})+(a_{1}+b_{1})a+...+(a_{n}+b_{n})a^{n}+...
= (a_{0} + a_{1}a + ... + a_{n}a^{n} + ...)
+(b_{0} + b_{1}a + ... + b_{n}a^{n} + ...)
= _{a}(f) + _{a}(g).
Similarly, since
f · g = c_{0} + c_{1}X + ... + c_{n}X^{n} + ...,
where
c_{i} = a_{0}b_{i} + a_{1}b_{i1} + ... +a_{i}b_{0},
we see that _{a}(f · g) = c_{0} + c_{1}a + c_{2}a^{2} + ... + c_{n}a^{n} + ... = (a_{0} + a_{1}a + ...)·(b_{0} + b_{1}a + ...) (R is commutative) =_{a}(f ) · _{a}(g). Therefore, _{a} is a ring homomorphism. We usually refer to _{a} as "evaluation at a." In this demonstration, let F be a field, X an indeterminate over F, Clearly
f = (X  x)q + r such that deg(r) < deg(q)
since x is a zero of f 0 = (Xx)q + r o this implies that either
(X  x)q = r or that r = 0 since deg(r) < deg(q) thus (Xx)q cannot equal r so r = 0 which means f = (X  x)q.

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