Question

whats the product of 8Square root 16D times 2d square root of 3? simplify if possible

Answer 1

Is that:
\(8\sqrt{16d}\cdot 2d \sqrt{3}\)

Answer 2

Radicals (roots) work a lot like fractions. To add two fractions, they must have a common denominator, but to multiply them you just multiply the top with the top and the bottom with the bottom. In radicals,.if you want to add them, the part under the radical must be the same. If you want to multiply them, you just do so.

\(\sqrt{5}+\sqrt{4}\) can not be added because the radicands are different. But \(\sqrt{5}+\sqrt{5}=2\sqrt{5}\)

\(\sqrt{5}\cdot \sqrt{4} = \sqrt{20} = 2\sqrt{5} = \sqrt{5}\cdot 2\) It works if I multiply first, then take out the root of 4, or if I do that last.

Answer 3

Now apply this to your problem: \(8\sqrt{16d}\cdot 2d \sqrt{3}\) You can simplify that first because of the 16 under the one root.
\((8\cdot 4)\sqrt{d}\cdot 2d \sqrt{3}\implies 32\sqrt{d}\cdot 2d \sqrt{3}\)
Then you would just multiply the rest.

Nothing really special about radical multiplication. Just radical addition.