1Cademy - Simplifying $$\sqrt{2} \cdot \sqrt{6}$$ and $$(4\sqrt{3})(2\sqrt{12})$$

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Product Property of Square Roots

Example

Simplifying $\sqrt{2} \cdot \sqrt{6}$ and $(4\sqrt{3})(2\sqrt{12})$

Multiply and simplify two products of square roots — one without coefficients and one with coefficients.

ⓐ $\sqrt{2} \cdot \sqrt{6}$ :

Apply the Product Property to combine the two radicals under a single radical sign:

$\sqrt{2} \cdot \sqrt{6} = \sqrt{12}$

The radicand $12$ is not a perfect square, so extract the largest perfect-square factor: $12 = 4 \cdot 3$ .

$\sqrt{12} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}$

The simplified result is $2\sqrt{3}$ .

ⓑ $(4\sqrt{3})(2\sqrt{12})$ :

Multiply the coefficients together and multiply the radicands together:

$(4\sqrt{3})(2\sqrt{12}) = 8\sqrt{36}$

Since $36$ is a perfect square, $\sqrt{36} = 6$ :

$8\sqrt{36} = 8 \cdot 6 = 48$

The result is 48. In part ⓑ, the radical $\sqrt{12}$ was not simplified before multiplying — waiting to multiply first produced $\sqrt{36}$ , a perfect square, which made the simplification straightforward.

Updated 2026-04-21

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OpenStax Elementary Algebra

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