So this is the formula I'm working with:
$$
E^2 = m^2c^4+p^2c^2
$$
From that we can get this:
$$
E = \sqrt{m^2c^4+p^2c^2}
$$
But I'm wondering what the process between ^ this equation and the one below is called.
Also could someone explain to me the reasoning behind this factorisation(?)
$$
E = \sqrt{m^2c^4(1+\frac{p^2c^2}{m^2c^4})}
$$
On
I'm not certain about a special name for the process indicated but the reason it works if because of factorization rules. If you factor a quantity out of another quantity, then you divide by the quantity to find the resultant term inside the expression.
The probable reason it's being done here is to compare the modified expression to the original. By factoring out $mc^2$, you can understand it relative to the original expression.
On
The reason for this factorization is physics driven. The main purpose is bring the expression to the form $$E = mc^2 \sqrt{1 + \frac{p^2}{m^2c^2}}$$ In the limit where $p \ll mc$, we use the fact $$\sqrt{1 + x} \approx 1 + \frac{x}{2}\quad\text{ for }\quad |x| \ll 1$$ to deduce
$$E \approx mc^2 \left( 1 + \frac{p^2}{2m^2c^2}\right) = mc^2 + \frac{p^2}{2m}$$
Translate to physical terms, it means when the momentum $p$ of an object is small, its energy is roughly the classical kinetic energy $\displaystyle\;\frac{p^2}{2m}$ plus a constant.
It is a simple use of the distributive property (Wikipedia link): $$\Large xy+xz=x(y+z)$$ In your situation, $x=m^2c^4$, $\,y=1$, and $\,z=\dfrac{p^2c^2}{m^2c^4}$.
As a commenter points out above, the purpose of this manipulation of the formula is likely to get $$E = \sqrt{m^2c^4\left(1+\frac{p^2c^2}{m^2c^4}\right)}=\sqrt{m^2c^4\Large\strut}\cdot\sqrt{1+\frac{p^2c^2}{m^2c^4}}=mc^2 \sqrt{ 1 + \frac{p^2c^2}{m^2 c^4} }=mc^2 \sqrt{ 1 + \frac{p^2}{m^2 c^2} }$$ which isolates $\sqrt{1+\frac{p^2}{m^2c^2}}$ as a sort of relativistic correction factor for the equation $E=mc^2$.