Trying to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ into $\frac{-5\sqrt{2}-6}{7}$ I'm asked to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ and am provided with the solution $\frac{-5\sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2\sqrt{2}$ thus:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$ = $\frac{2\sqrt{2}-4}{4-\sqrt{2}} * \frac{4+\sqrt{2}}{4+\sqrt{2}}$ = $\frac{(2\sqrt{2}-4)(4+\sqrt{2})}{16-2}$ = 
(I become less certain in my working here)
$\frac{8\sqrt{2}*2(\sqrt{2}^2)-16-4\sqrt{2}}{14}$ = $\frac{8\sqrt{2}*4-16-4\sqrt{2}}{14}$ = $\frac{32\sqrt{2}-16-4\sqrt{2}}{14}$ = $\frac{28\sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $\sqrt{2}$
This is as far as I can get. I don't know if $\frac{28\sqrt{2}-16}{14}-\sqrt{2}$ is still correct or close to the solution. How can I arrive at $\frac{-5\sqrt{2}-6}{7}$?
 A: $$\begin{align}
\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-\sqrt{2}&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\sqrt{2}\\
&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\frac{4\sqrt{2}-2}{4-\sqrt{2}}\\
&=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}\\
&=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}~\cdot~\frac{4+\sqrt{2}}{4+\sqrt{2}}\\
&=\frac{-10\sqrt{2}-12}{14}\\
&=\frac{-5\sqrt{2}-6}{7}
\end{align}$$
A: You were doing fine until the place where you tried to expand 
$(2\sqrt2 - 4)(4 + \sqrt2).$
There are mnemonic techniques for this but I think plain old distributive law works well enough:
\begin{align}
(2\sqrt2 - 4)(4 + \sqrt2) &= (2\sqrt2 - 4)4  + (2\sqrt2 - 4)\sqrt2 \\
 &= (8\sqrt2 - 16) + (4 - 4\sqrt2) \\
 &= 4\sqrt2 - 12.
\end{align}
Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of
$\frac{4\sqrt2 - 12}{14}.$
And finally you'll want to change the $-\sqrt2$ so that you have two fractions with a common denominator and can finish.
A: \begin{align}
\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2} & = \frac{2\sqrt{2}-4}{4-\sqrt{2}}\cdot \frac{4+\sqrt{2}}{4+\sqrt{2}} - \sqrt{2}  \\
& = \frac{4\sqrt{2}-12}{14} - \sqrt{2} \\
& = \frac{2\sqrt{2}-6}{7} - \sqrt{2} \\
& = \frac{2\sqrt{2}-6 -7 \sqrt{2}}{7}\\
& = \frac{-5\sqrt{2} -6 }{7}
\end{align} 
