Solve (sqrt{6}-4sqrt{2})(3sqrt{6}+sqrt{2})

$$( \sqrt { 6 } - 4 \sqrt { 2 } ) ( 3 \sqrt { 6 } + \sqrt { 2 } )$$

Evaluate

$$10-22\sqrt{3}\approx -28.105117767$$

Solution Steps

Apply the distributive property by multiplying each term of \sqrt{6}-4\sqrt{2} by each term of 3\sqrt{6}+\sqrt{2}.

$$3\left(\sqrt{6}\right)^{2}+\sqrt{6}\sqrt{2}-12\sqrt{2}\sqrt{6}-4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{6} is 6.

$$3\times 6+\sqrt{6}\sqrt{2}-12\sqrt{2}\sqrt{6}-4\left(\sqrt{2}\right)^{2}$$

Multiply 3 and 6 to get 18.

$$18+\sqrt{6}\sqrt{2}-12\sqrt{2}\sqrt{6}-4\left(\sqrt{2}\right)^{2}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$18+\sqrt{2}\sqrt{3}\sqrt{2}-12\sqrt{2}\sqrt{6}-4\left(\sqrt{2}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$18+2\sqrt{3}-12\sqrt{2}\sqrt{6}-4\left(\sqrt{2}\right)^{2}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$18+2\sqrt{3}-12\sqrt{2}\sqrt{2}\sqrt{3}-4\left(\sqrt{2}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$18+2\sqrt{3}-12\times 2\sqrt{3}-4\left(\sqrt{2}\right)^{2}$$

Multiply -12 and 2 to get -24.

$$18+2\sqrt{3}-24\sqrt{3}-4\left(\sqrt{2}\right)^{2}$$

Combine 2\sqrt{3} and -24\sqrt{3} to get -22\sqrt{3}.

$$18-22\sqrt{3}-4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$18-22\sqrt{3}-4\times 2$$

Multiply -4 and 2 to get -8.

$$18-22\sqrt{3}-8$$

Subtract 8 from 18 to get 10.

$$10-22\sqrt{3}$$