Solve (3sqrt{2}+8sqrt{3})(4sqrt{2}-3sqrt{3})

$$( 3 \sqrt { 2 } + 8 \sqrt { 3 } ) ( 4 \sqrt { 2 } - 3 \sqrt { 3 } )$$

Evaluate

$$23\sqrt{6}-48\approx 8.338264084$$

Solution Steps

Apply the distributive property by multiplying each term of 3\sqrt{2}+8\sqrt{3} by each term of 4\sqrt{2}-3\sqrt{3}.

$$12\left(\sqrt{2}\right)^{2}-9\sqrt{3}\sqrt{2}+32\sqrt{3}\sqrt{2}-24\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$12\times 2-9\sqrt{3}\sqrt{2}+32\sqrt{3}\sqrt{2}-24\left(\sqrt{3}\right)^{2}$$

Multiply 12 and 2 to get 24.

$$24-9\sqrt{3}\sqrt{2}+32\sqrt{3}\sqrt{2}-24\left(\sqrt{3}\right)^{2}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$24-9\sqrt{6}+32\sqrt{3}\sqrt{2}-24\left(\sqrt{3}\right)^{2}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$24-9\sqrt{6}+32\sqrt{6}-24\left(\sqrt{3}\right)^{2}$$

Combine -9\sqrt{6} and 32\sqrt{6} to get 23\sqrt{6}.

$$24+23\sqrt{6}-24\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$24+23\sqrt{6}-24\times 3$$

Multiply -24 and 3 to get -72.

$$24+23\sqrt{6}-72$$

Subtract 72 from 24 to get -48.

$$-48+23\sqrt{6}$$

Factor

$$23 \sqrt{6} - 48 = 8.338264084$$