$2x^2+13x+15=2x^2+10x+3x+15=2x(x+5)+3(x+5)=\frac{(2x+3)(x+5)}{(x+5)}=2x+3$

Substituting $x=2$ in resulting expression gives $2\times 2+3=7$.

$(a+b{)}^2+(a-b{)}^2=(a^2+b^2+2ab)+(a^2+b^2-2ab)=2(a^2+b^2)$
Dividing by $a^2+b^2$ we get,
= $\frac{2(a^2+b^2)}{(a^2+b^2)}=2$
Thus, $\frac{\left(\right(a+b{)}^2+(a-b{)}^2)}{(a^2+b^2)}=2$

$a^2-2ab+b^2-c^2=(a-b{)}^2–c^2$
Using identity $x^2-y^2=(x+y)(x-y)$  we get
$(a-b{)}^2–c^2$ = $\left[\right(a-b)+c]\left[\right(a-b)-c]$​
Hence, factors of $a^2-2ab+b^2-c^2$ are $a-b+c$ and $a-b-c$

Given, the expression is
$4x^3+12x^2+5x+15.$
Taking $4x^2$ common from the first two-terms and 5 common from the last two-terms, we get
$4x^2(x+3)+5(x+3)$
$=(x+3)(4x^2+5)$
$\text{Thus,}(x+3)\text{and}(4x^2+5)$
are two factors of the given expression.

$\text{Given:}\frac{a^2+7a+10}{(a+5)}...\left(i\right)$
Comparing $a^2+7a+10$ with the identity $x^2+(a+b)x+ab,$
$\text{we note that,}(a+b)=7andab=10$
$So,5+2=7and\left(5\right)\left(2\right)=10$
Hence,
$a^2+7a+10$
$=a^2+5a+2a+10$
$=a(a+5)+2(a+5)$
$=(a+2)(a+5)$
From (i), we get
$\frac{a^2+7a+10}{(a+5)}=\frac{(a+2)(a+5)}{(a+5)}$
$=(a+2)$

Given, the expression is
$\frac{m^2+7m-60}{(m-5)}.$
Now, comparing the numerator with the identity $x^2+(a+b)x+ab$
We note that,
$(a+b)=7andab=-60.$
So,
$12+(-5)=7and\left(12\right)(-5)=-60$.
Hence,
$m^2+7m-60=m^2+12m–5m–60=m(m+12)–5(m+12)=(m-5)(m+12)$
Therefore,
$\frac{m^2+7m-60}{(m-5)}=\frac{(m-5)(m+12)}{(m-5)}=(m+12)$ 

$\text{Given, the expression is}25+70x+49x^2.$
The given expression can be rewritten as,
$49x^2+70x+25$
$\text{Using the identity:}(a+b{)}^2=a^2+2ab+b^2,\text{we get}$
$49x^2+70x+25=(7x{)}^2+2(5\left)\right(7x)+5^2$
$=(7x+5{)}^2$​
$=(7x+5)(7x+5)$
$\therefore 25+70x+49x^2=(7x+5)(7x+5)$

The given expression can be written as,
$49p^2-36=(7p{)}^2-(6{)}^2$
$\text{[Using the identity:}{a}^2-b^2=(a+b)(a-b)]$
$=(7p+6)(7p-6)$
$\text{Hence,}(7p+6)and(7p-6)$
are the factors of $49p^2-36.$ 

Comparing the coefficients of the different elements and dividing, we get:

$\frac{24}{6}=4$

$\frac{x}{1}=x$

$\frac{y^2}{y}=y$

$\frac{z^2}{z^2}=1$

Hence, we have $24x{y}^2{z}^2÷6y{z}^2=4xy$.

Divide all the terms of the expression $(x^3+2x^2+3x)$ separately by $2x$. 
$\frac{x^3}{2x}=\frac{x^2}{2}$;
 $\frac{2x^2}{2x}=x$;
$\frac{3x}{2x}=\frac{3}{2}$.
Therefore, $\frac{(x^3+2x^2+3x)}{2x}=\frac{x^2}{2}+x+\frac{3}{2}=\frac{(x^2+2x+3)}{2}$