Prove that $\gcd(pf,pg) = p \cdot \gcd(f,g)$ when $p,f,g$ are polynomials How does one start proving this theory?
Prove:
$$\gcd(pf,pg) = p \cdot \gcd(f,g)$$
when
$$p,f,g \in \mathbb F[x] \;,\;\text{The max power multiplier of $p$ is 1 (fixed polynomial)}.$$
 A: Hint:
What's the GCD$(ak,bk)$, if GCD$(a,b)=1$?
For example: Let $a(x)=x^2+x+2$ and $b(x)=x^2+x+1$. Clearly, GCD$(a(x),b(x))=1$
If you multiply $a(x)$ and $b(x)$ with some polynomial $k(x)$, you will have GCD of new polynomials to be $k(x)$. 
A: Inceptio's hint will let you prove that equality whenever your polynomial ring is a UFD. If you assume that $\Bbb F$ is a field, though, it may be easier to recall that $\Bbb F[X]$ is a PID...
A: Below are three proofs of the gcd distributive law $\rm\ (ax,bx) = (a,b)x\ $  using Bezout's identity,  universal gcd laws, and unique factorization.

First we show that the gcd distributive law follows immediately from the fact that, by Bezout, the gcd may be specified by linear equations. Distributivity follows because such linear equations are preserved by scalings. Namely, for  $\rm\:a,b,c,x \ne 0\:$ in a domain $\rm\:D\:$ satisfying Bezout's identity
$\rm\qquad\qquad \phantom{ \iff }\ \ \ \:\! c = (a,b) $
$\rm\qquad\qquad \iff\ \: c\:\ |\ \:a,\:b\ \ \ \  \ \ \&\ \ \ \ c\ =\ ja\: +\: kb,\ \  \ $ some $\rm\:j,k\in D$
$\rm\qquad\qquad \iff\  cx\ |\ ax,bx\ \ \ \&\ \ \ cx = jax + kbx,\ \,$ some $\rm\:j,k\in D$
$\rm\qquad\qquad { \iff }\ \   cx = (ax,bx) $
The reader familiar with ideals will note that these equivalences are captured more concisely in the distributive law for ideal multiplication $\rm\:(a,b)(x) = (ax,bx),\:$ when interpreted in a PID or Bezout domain, where the ideal $\rm\:(a,b) = (c)\iff c = gcd(a,b)$

Alternatively, more generally, in any integral domain $\rm\:D\:$ we have
Theorem $\rm\ \ (a,b)\ =\ (ax,bx)/x\ \ $ if $\rm\ (ax,bx)\ $ exists in $\rm\:D.$
Proof $\rm\quad\: c\ |\ a,b \iff cx\ |\ ax,bx \iff cx\ |\ (ax,bx) \iff c\ |\ (ax,bx)/x$ 
The above proof uses the universal definitions of GCD, LCM, which often serve to simplify proofs, e.g. see this proof of the GCD * LCM law.

Alternatively, comparing powers of primes in unique  factorizations, it reduces to the following
$$ \min(a+c,\,b+c)\ =\ \min(a,b) + c$$
The proof is precisely the same as the prior proof, replacing gcd by min, and divides by $\le$, and
$$\begin{eqnarray} {\rm employing}\quad\ c\le a,b&\iff& c\le \min(a,b),\\
\rm the\ analog\ of\quad\   c\  \, |\, \ a,b&\iff&\rm c\ \,|\,\ (a,b) \end{eqnarray}$$
