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Does anyone have any ideas about this one? | Does anyone have any ideas about this one? | ||
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| + | Try looking at the proof of Theorem 17.2. -Kristie | ||
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| + | Let r = p/q and f(x) is primitive | ||
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| + | then h(x) = f(x) / (x + p/q) | ||
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| + | Note that h(x) is also in Z[x] | ||
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| + | f(x) = h(x) * (x + p/q) | ||
| + | q* f(x) = h(x) * (q*x + p) | ||
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| + | 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. | ||
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| + | ---- | ||
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| + | The above proof doesn't really make sense to me, so here is what I used. | ||
| + | if | ||
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| + | <math>f(x) = x^{n} + a_{n-1}*x^{n-1} + ... + a_{0}</math> | ||
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| + | and g(x) = x-r is a factor of f(x), then | ||
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| + | <math> f(x) = (x-r) * ( x^{n-1} + a_{n-2}*x^{n-2} + ... + a_{0} ) </math> | ||
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| + | 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): | ||
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| + | <math> f(x) = (bx-a) * ( d * x^{n-1} + d * a_{n-2}*x^{n-2} + ... + d * a_{0} ) </math> | ||
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| + | But since the coefficient of x^n = 1 in f(x), b and d must be one. Therefore r = a / 1 = a, an integer. | ||
Latest revision as of 16:49, 16 November 2008
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.
