algebra precalculus - Why is $frac{sqrt{32}}{sqrt{14D}}=frac{4sqrt{7D}}{7D}$ and not $2sqrt{8}$?

I am asked to simplify $\frac{\sqrt{32}}{\sqrt{14D}}$ and am provided with the solution $\frac{4\sqrt{7D}}{7D}$.

(Side question, in the solution why can't $\sqrt{7D}$ and $7D$ cancel out since one is in he denominator so if multiplying out would it not be $\sqrt{7D} * 7D = 0$?)

I gave this question a try but arrived at $2\sqrt{8}$. Here's my working:

(multiply out the radicals in the denominator)

$\frac{\sqrt{32}}{\sqrt{14D}}$ =

$\frac{\sqrt{32}}{\sqrt{14D}} * \frac{\sqrt{14D}}{\sqrt{14D}}$ =

$\frac{\sqrt{32}\sqrt{14D}}{14D}$ =

(Use product rule to split out 32 in numerator)

$\frac{\sqrt{4}\sqrt{8}\sqrt{14D}}{14D}$ =

$\frac{2\sqrt{8}\sqrt{14D}}{14D}$ =

Then, using my presumably flawed logic in my side question above I cancelled out $14D$ to arrive at $2\sqrt{8}$

Where did I go wrong and how can I arrive at $\frac{4\sqrt{7D}}{7D}$?

Answer

Hint: Multiplying numerator and denomninator by $$\sqrt{14D}$$ we get $$\frac{\sqrt{32}\sqrt{14D}}{14D}$$ and this is equal b $$\frac{4\sqrt{2}{\sqrt{2}}\sqrt{7D}}{14D}$$ and this is equal to $$\frac{4\sqrt{7D}}{7D}$$ for $$D\neq 0$$
$$\sqrt{32}=\sqrt{2\cdot 16}=4\sqrt{2}$$ and $$\sqrt{14}=\sqrt{2}\sqrt{7}$$