How to solve the following arithmetic radical problem? $$ 2(4\sqrt{7} + 1 + 3\sqrt{7} + 2) $$
I distribute first right? 
$$ 8\sqrt{14} + 2 + 6\sqrt{14} + 4$$
$$ 14\sqrt{14} + 6$$
BUT IT LOOKS LIKE ITS SUPPOSED TO BE 
$$14\sqrt{7} + 6$$
I also have a question if you have to multiply $$ 2(5\sqrt{5})$$ do you multiply the 5 inside the √? I forgot about these. What were the general rules for these?
 A: No, you don't multiply inside the radical. Forget about $\sqrt{7}$ for the time being. Let's redistribute $$2(4x + 1 + 3x + 2)$$ So at this point we choose not to know what $x$ is. Imagine you have four boxes. They all look identical, you don't know what their contents are, but you do know they all contain the same thing. If you add four of those boxes to another three of those boxes you then have seven boxes. Thus $$2(3 + 7x)$$ which we then redistribute to get $$6 + 14x.$$ Now let's open the boxes to find out that $x = \sqrt{7}$ and therefore our redistribution is $$6 + 14 \sqrt{7}$$ (we're dealing with commutative algebra, so this is the same thing as $14 \sqrt{7} + 6$).

It's the same for your parallel question: $2(5 \sqrt{5}) = 10 \sqrt{5}$, not $10 \sqrt{10}$. Verify this on a calculator: $\sqrt{5} \approx 2.236067977$. Then 2 * 5 * 2.236067977 = 22.36067977, that is, ten times the square root of 5.
A: For non-negative real numbers $a$ and $b$, we have $$a\sqrt{b} = \sqrt{a^2 b}.$$
However, this is not necessary to simplify your expression, 
since $$a(b\sqrt{c}) = (ab)\sqrt{c} \ne ab\sqrt{ac},$$
the latter being what you have repeatedly done.
A: Recall that $a(bc)=(ab)c$.  Thus, $2(4\sqrt{7})=(2\cdot 4)\sqrt{7}$.
