Divide inside a Radical It has been so long since I have done division inside of radicals that I totally forget the "special rule" for for doing it. -_-
For example, say I wanted to divide the 4 out of this expression:

$\sqrt{1 - 4x^2}$

Is this the right way to go about it?

$\frac{16}{16} \cdot \sqrt{1 - 4x^2}$
$16 \cdot \frac{\sqrt{1 - 4x^2}}{16}$
$16 \cdot \sqrt{\frac{1 - 4x^2}{4}} \Longleftarrow \text{Took the square root of 16 to get it in the radicand as the divisor}$

I know that this really a simple, question. Can't believe that I forgot how to do it. :(
 A: There is, in fact, one rather important mistake there.  16 is not the square root of 4.  If you replace '16' by'2' in your equation then it is all right.  In general, if you divide by everything inside the square root sign by some constant then you should multiply in front by the square root of that constant.   
A: Square roots obey the rule $\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}$ .  You wanted to take the $4$ out of $\sqrt{1-4x^2}$.  $1-4x^2=4\cdot\frac{1-4x^2}{4}$.  So $\sqrt{1-4x^2}=\sqrt{4\cdot\frac{1-4x^2}{4}}=\sqrt{4}\cdot\sqrt{\frac{1-4x^2}{4}}=2\cdot\sqrt{\frac{1-4x^2}{4}}$.  I think that's the best explanation I can give.
A: The correct way to do this, after fixing the mistake pointed out by Donkey_2009, is:

$\dfrac{2}{2} \cdot \sqrt{1-4x^2}$
$= 2 \cdot \dfrac{\sqrt{1-4x^2}}{2}$
$= 2 \cdot \dfrac{\sqrt{1-4x^2}}{\sqrt{4}} \qquad \Leftarrow$ applied $x = \sqrt{x^2}$
$= 2 \cdot \sqrt{\dfrac{1-4x^2}{4}} \qquad \Leftarrow$ applied $\frac{\sqrt a}{\sqrt b} = \sqrt{\frac a b}$

A: First divide everything inside the radical by what you will want to take out of it, factoring this out while keeping it inside the radical. Then square root what you want to take out of the radical, and take it out. 
