A polynomial with integer coefficients is called primitive if its coefficients are relatively prime. For example, $$3{x^2} + 7x + 9$$ is primitive while $$10{x^2} + 5x + 15$$ is not.
(a) Prove that the product of two primitive polynomials is primitive.
(b) Use this to prove Gauss's Lemma: If a polynomial with integer coefficients can be factored into polynomials with rational coefficients, it can also be factored into primitive polynomials with integer coefficients
On
Since you tagged it elementary number theory and not abstract algebra, here is an approach that does not require any knowledge of ring theory.
Write the polynomials as $\,A,B.\,$ Suppose $\,p\nmid A,B.\,$ So when reduced mod $\,p,\,$ both are $\not\equiv 0$ hence they have leading coef's $\,a,b\not\equiv 0\pmod p.\,$ By $\,p\,$ prime: $\,p\nmid a,b\,\Rightarrow\, \color{#c00}{p\nmid ab},\,$ so
$$ \begin{eqnarray} A &\equiv&\ \ a\ x^j + \:\cdots,\quad\ \ a\not\equiv 0\pmod p\\ B &\equiv&\ \ b\ x^k + \:\cdots,\quad\ \ b\not\equiv 0\pmod p\\ \Rightarrow\ AB &\equiv& \color{#c00}{ab}\ x^{j+k}\! + \:\cdots,\ \color{#c00}{ab\not\equiv 0}\pmod p\end{eqnarray}$$
So $\,p\nmid AB.\,$ Said contrapositively, $\ p\mid AB\,\Rightarrow\,p\mid A\,$ or $\,p\mid B,\, $ i.e. a prime $\,p\in \Bbb Z\,$ remains prime in $\,\Bbb Z[x] = $ polynomials with integer coefficients. This preservation of the prime divisor property in the leading coef's will become clearer when one studies ring theory, where it follows from general structural results [$\,p$ prime $\Rightarrow \Bbb Z_p$ domain $\Rightarrow$ $\,\Bbb Z_p[x]$ domain].
For part 2, factor out fractional contents, then clear denominators and apply the first part.
Remark $ $ Note that when reduced mod $p$ the polynomial may have different leading coef, e.g. $\, A = 10x^2+ax\equiv ax\pmod{5}.\,$ But $\,p\nmid A\,\Rightarrow\,A\not\equiv 0\,$ so its reduction has lead coef $\,a\not\equiv 0$
For the first part let $f$ and $g$ be primitive polynomials in $\mathbb{Z}$ and assume their product isn't, i.e. there exists prime number $p$ that divides all coefficients of $f\cdot g$. Now define the following mapping:
$$\phi = \begin{cases} \mathbb{Z}[x] \to \mathbb{Z}_p[x] \\ a_nx^n + a_{n-1}x^{n-1} + ... a_0 \to [a_n]x^n + [a_{n-1}]x^{n-1} + ... + [a_0] \end{cases}$$
Prove that it's epimorphism and as $\mathbb{Z}_p[x]$ is an integral domain we have that $\phi(f)\phi(g) = \phi(f \cdot g) = 0$ implies that one of the polynomials isn't primitive.
For the second part let $f,g \in \mathbb{Q}[x]$. Let $b$ be the LCM of the denominators and $a$ be the GCD of the nominators of $f$ and similarly define $c,d$ for $g$. Now $f = \frac{a}{b} \cdot F$ and $g = \frac{c}{d} \cdot G$, where $F,G \in \mathbb{Z}[x]$ and they are primitive.
Then we have $h = \frac{a}{b} \cdot F \cdot \frac{c}{d} \cdot G \implies bd \cdot h = ac \cdot F \cdot G$. But $F\cdot G$ is primitive, as well as $h$, so we must have $ac = bd$. So: $$h = f \cdot g = F \cdot G$$
Here's a more elementary solution to the first part. Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ and $g(x) = b_tx^t + b_{t-1}x^{t-1} + ... + b_0$. Now assume that $p \mid f \cdot g$ for some prime number $p$. As both are primitive there exist coefficients $a_i$ and $b_j$ that aren't divisible by $p$. Let $a_i$ and $b_j$ be the coefficients such that $p \not \mid a_i, b_j$, but $p \mid a_s, b_u$ for $s>i$ and $u>j$. Now let's consider the coefficient in front of $x^{i+j}$ in $f \cdot g$. We have:
$$ c_{i+j} = a_0b_{i+j} + a_1b_{i+j-1} + ... + a_ib_j + ... + a_{i+j-1}b_1 + a_ib_0$$
Using the previously mentioned things we have that $0 \equiv c_{i+j} \equiv a_ib_j \pmod p$, an ovcious contradiction. Hence the claim is proven by contradiction.