Using the binomial expansion to approximate a square root. 
Find the first four terms of the expansion $\sqrt{1-4x}$ in ascending powers of x.
  Hence, approximate $\sqrt{6}$ to four decimal places.

I've expanded it to form this: $1-2x-2x^2-4x^3$
When approximating the value of $\sqrt{6}$, bearing in mind that for the expansion to be valid , $\left|x\right|<\frac{1}{4}$, I took out a value of $\sqrt{10}$ to form $\sqrt{10}\sqrt{0.6}$. Then, letting $x = 0.1$, I substituted $x$ into the expansion, then multiplied by $\sqrt{10}$ to find an approximate value for $\sqrt{6}$. Is there anything invalid about this? Why is it better to take out a factor of $2.5$?
 A: Here we look at some arguments why taking out a    factor   $2.5$ is a convenient choice.

Problem: We want to (manually)  approximate $\sqrt{6}$ by using the first few terms of the binomial series expansion of 
  \begin{align*}
\sqrt{1-4x}&=  \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-4x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{4}\\
&= 1-2x-2x^2-4x^3+\cdots\tag{1}
\end{align*}

In order to apply (1) we are  looking for a number $y$ with
\begin{align*}
\sqrt{1-4x}&=\sqrt{6y^2}=y\sqrt{6}\tag{2}\\
\color{blue}{\sqrt{6}}&\color{blue}{=\frac{1}{y}\sqrt{1-4x}}
\end{align*}
We see it is convenient to choose $y$ to be a square number which can be easily factored out from the root.  We obtain from (2)
\begin{align*}
1-4x&=6y^2\\
4x&=1-6y^2\\
\color{blue}{x}&\color{blue}{=\frac{1}{4}-\frac{3}{2}y^2}\tag{3}
\end{align*}

When looking for a nice $y$  which fulfills (3) there are some aspects to consider:
  
  
*
  
*We  have  to respect the radius of convergence $|x|<\frac{1}{4}$.
  
*Since we want to calculate an approximation of $\sqrt{6}$ by hand we should take $y\in\mathbb{Q}$ with rather small numbers as numerator and denominator.
  
*Last but not least: We want to find a value $x$ which provides a good approximation for $\sqrt{6}$.
We will see it's not hard to find values which have these properties.

We see in (1) a good approximation is given if $x$ is close to $0$. If $x$ is close to zero we will also fulfill the convergence condition. $x$ close to zero means that in (3) we have to choose $y$ so that
\begin{align*}
\frac{3}{2}y^2
\end{align*} 
is close to $\frac{1}{4}$. So,  $y^2$ is close to $\frac{1}{4}\cdot\frac{2}{3}=\frac{1}{6}$. We have already   (1)  and (3) appropriately considered. Now we want to find  small natural  numbers $a,b$ so that
\begin{align*}
y^2=\frac{a^2}{b^2}\approx \frac{1}{6}
\end{align*}
This can be done easily. When going through small numbers of $a$ and $b$ whose squares are apart by a factor $6$ we might quickly come to $100$ and $16$. These are two small squares and we have $6\cdot 16=96$ close to $100$. That's all.

Now it's time to harvest. We choose $y^2=\frac{16}{100}=\frac{4}{25}$ resp. $y=\frac{2}{5}$. We obtain for $x$ from (3)
  \begin{align*}
x=\frac{1}{4}-\frac{3}{2}y^2=\frac{1}{4}-\frac{3}{2}\cdot\frac{4}{25}=\frac{1}{100}
\end{align*}
  We have now an appropriate value $x=\frac{1}{100}$ and we finally get  from (1) the approximation:
  \begin{align*}
\color{blue}{\sqrt{6}} \approx \frac{5}{2}\left(1-2\cdot 10^{-2}-2\cdot 10^{-4}- 4\cdot 10^{-4}\right)\color{blue}{=2.449\,4}9
\end{align*}
We have $\sqrt{6}=\color{blue}{2.449\,4}89\,7\ldots$ with an approximation error $\approx 2.572\times 10^{-7}$. This result is quite impressive when considering that we have used just four terms of the binomial series.

At this point it might be clear that factoring out $2.5=\frac{5}{2}=\frac{1}{y}$ is a good choice.
Note: In a section about binomial series expansion in Journey through Genius by W. Dunham the author cites Newton: Extraction of roots are much shortened by this theorem, indicating how valuable this technique was for Newton.
