Solve: Product of Square Roots (√2 × √6 × √12) ÷ √16

Yes! This is exactly what the multiplication property tells us: $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$. So $\sqrt{2} \cdot \sqrt{6} \cdot \sqrt{12} = \sqrt{2 \cdot 6 \cdot 12}$.

Try thinking of common perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. Since $12 \times 12 = 144$, we know $\sqrt{144} = 12$!

You could, but it's more work! $\sqrt{12} = 2\sqrt{3}$ creates messier calculations. It's easier to use the multiplication property and combine everything under one radical first.

Since $\sqrt{16} = 4$ (a whole number), there's no radical left in the denominator to rationalize. The final answer 3 is already in simplest form.

They're the same thing! The multiplication property of radicals says $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$. We can work in either direction depending on what makes the problem easier.