Let F be a field and let f(x), g(x), h(x), and d(x) be polynomials in F[x]. Prove that if d(x)=gcd(f(x),g(x)) and both f(x) and g(x) divide h(x), then f(x)g(x) divides h(x)d(x).

To prove that f(x)g(x) divides h(x)d(x), we need to show that there exists a polynomial k(x) such that h(x)d(x) = f(x)g(x)k(x). We can do this by applying polynomial division.

1. Let's start by using the fact that both f(x) and g(x) divide h(x). This means that there exist polynomials p(x) and q(x) such that h(x) = f(x)p(x) and h(x) = g(x)q(x).

2. Since h(x) = f(x)p(x) = g(x)q(x), we can equate the two expressions and get f(x)p(x) = g(x)q(x).

3. Now, let's consider the polynomial d(x) = gcd(f(x), g(x)). By definition of greatest common divisor, d(x) divides both f(x) and g(x). Hence, there exist polynomials r(x) and s(x) such that f(x) = d(x)r(x) and g(x) = d(x)s(x).

4. Substituting these expressions into the equation from step 2, we have (d(x)r(x))p(x) = (d(x)s(x))q(x).

5. Rearranging the terms, we get d(x)(r(x)p(x) - s(x)q(x)) = 0.

6. Since F is a field, F[x] is an integral domain, meaning that the only way a product of polynomials can be zero is if at least one of the polynomials is zero.

7. Therefore, we have two possibilities: either d(x) = 0 or r(x)p(x) - s(x)q(x) = 0.

8. The first possibility, d(x) = 0, contradicts the assumption that d(x) is the greatest common divisor of f(x) and g(x). Hence, we can discard it.

9. This leaves us with the second possibility: r(x)p(x) - s(x)q(x) = 0.

10. Now, let's multiply both sides of this equation by f(x) and g(x): f(x)r(x)p(x) - g(x)s(x)q(x) = 0.

11. Simplifying, we obtain f(x)g(x)(r(x)p(x)) = g(x)s(x)q(x).

12. This implies that f(x)g(x) divides h(x)d(x), where k(x) = r(x)p(x).

Therefore, we have shown that if d(x) = gcd(f(x), g(x)) and both f(x) and g(x) divide h(x), then f(x)g(x) divides h(x)d(x).