$14pq$ itself is a factor of $28p^{2}q$. Thus, any factor of $14pq$ will also be a factor of $28p^{2}q$.

Given,
$\frac{x(x+1)(x+2)(x+3)}{x(x+1)}$
On cancelling the common terms $xand(x+1)$ , we get 
$\Rightarrow (x+2)(x+3)$

$2t,3t^2\text{and}4\text{can be factorized as,}$
$2t=1\times 2\times t$
$3t^2=1\times 3\times t\times t$
$4=1\times 2\times 2$ 
Hence, 1 is the only common factor among all three.

The given expression can be rewritten as,
$(2y{)}^2-2(2y\left)\right(3)+3^2.$
Comparing with the identity
$(a-b{)}^2=a^2-2ab+b^2$
we get,
$(2y{)}^2-2(2y\left)\right(3)+3^2$
$=(2y-3{)}^2$
$=(2y-3)(2y-3)$

We need to regroup ax  + bx - ay - by

We get x(a+b) - y(a+b)

Taking (a+b) common from both the terms we get (a+b)(x-y)
Thus, (a+b) and (x-y) is the factors of  ax  + bx - ay - by

$4y(y+3)$ can be written as
$4\times y\times (y+3)$ 
$=4y\times (y+3)=4(y+3)\times y$
$=4\times y(y+3)$ 
$=4\times (y^2+3y)$
$=4y^2+12y$ 
So, 4$y^2$ is a term in the expansion of $4y(y+3)$. It is not a factor.

$4x$($x+3$) = $4x^2$ + 3
So, $4x^2$ is a term of a given expression. $4x^2$ + 3 is not divisible by $4x^2$.
So, $4x^2$ is not a factor of $4x^2+3$.
The factors of  $4x$($x+3$) are 
$4,xand(x+3)$

Given, the expression is
$3m^2+9m+6.$
Taking 3 common from the above expression, we get
$3(m^2+3m+2)...\left(i\right)$
Now comparing $m^2+3m+2$ with the identity $x^2+(a+b)x+ab.$
$\text{We note that,}$
$(a+b)=3andab=2.$
$So,$
$2+1=3and\left(2\right)\left(1\right)=2$
Hence,
$m^2+3m+2$
$=m^2+2m+m+2$
$=m(m+2)+1(m+2)$
$=(m+2)(m+1)$
Now from (i), we get
$\Rightarrow 3m^2+9m+6=3(m^2+3m+2)=3(m+2)(m+1)$
Therefore,
3, (m+2) and (m+1) are the 3 irreducible factors of the given expression.

The given expression is
$x^2-2x-15$
We have to split the middle term in such a way the sum is -2 and product is -15 , which can be done in following way
$=x^2-5x+3x-15=x(x-5)+3(x-5)$
$=(x+3)(x-5)$

So, $x+3$ is one of the factors of the given expression.

Since 14 is HCF of 4p and 21q, then both of them should have 7 and 2 as prime factors.
4p = 2 x 2 x p 
21q = 3 x 7 x q
Since 14 is the HCF, p should be 7 and q should be 2.