Does anyone have any ideas about this one?
Try looking at the proof of Theorem 17.2. -Kristie
Let r = p/q and f(x) is primitive
then h(x) = f(x) / (x + p/q)
Note that h(x) is also in Z[x]
f(x) = h(x) * (x + p/q) q* f(x) = h(x) * (q*x + p)
The left side is not primitive therefore it is not in Z[x] while the right still is because p and q are relatively prime.
The above proof doesn't really make sense to me, so here is what I used. if
$ f(x) = x^{n} + a_{n-1}*x^{n-1} + ... + a_{0} $
and g(x) = x-r is a factor of f(x), then
$ f(x) = (x-r) * ( x^{n-1} + a_{n-2}*x^{n-2} + ... + a_{0} ) $
Let r = a / b for a,b in Z. We can multiply by b (the lcm of the denominators of g(x)) and some number d (the lcm of the denominators of h(x)), to get two new functions which multiply to give f(x):
$ f(x) = (bx-a) * ( d * x^{n-1} + d * a_{n-2}*x^{n-2} + ... + d * a_{0} ) $
But since the coefficient of x^n = 1 in f(x), b and d must be one. Therefore r = a / 1 = a, an integer.
