Abstract Algebra: Let f, g, h be elts of F[x] w/ f !=0. If

[teachhs]

Here is my problem: Let f(x), g(x), h(x) be elements of F[x] such that f(x) does not equal 0. Prove that: If f(x) divides g(x) and f(x) divides [(2x<sup>2 - 7x)g(x) - h(x)], then f(x) divides h(x).

So far I have:

Proof. Assume f(x) divides g(x) and g(x) divides h(x). Then there is s(x), t(x) are elements of F[x] such that g(x) = f(x)s(x) and h(x) = g(x)t(x).

This is where I am stumped. I know that

If f(x) divides g(x) and g(x) divides h(x) then f(x) divides h(x), and If f(x) divides g(x) and f(x) divdies (g(x) + h(x)), then f(x) divides h(x),

but how do I figure out my proof?</sup>

 

[stapel]

Proof. Assume...g(x) divides h(x).

On what basis are you making this assumption? Where do you use the fact that h(x) = (2x[sup:1qwuk2p1]2[/sup:1qwuk2p1] - 7x)(g(x))? (And what is the meaning of "F[x]"?)

Thank you!

Eliz.

 

[daon]

This is true in any ring... Its only a slight modification of the flowing proposition:

If a|b and a|(b+c) then a|c.

(1) b = ar
(2) (b+c) = as

(2) => c = (as-b) = (as-ar) = a(s-r)