Solve (4sqrt{2}-2sqrt{3})(2sqrt{3}+5sqrt{2})

$$( 4 \sqrt { 2 } - 2 \sqrt { 3 } ) ( 2 \sqrt { 3 } + 5 \sqrt { 2 } )$$

Evaluate

$$28-2\sqrt{6}\approx 23.101020514$$

Solution Steps

Apply the distributive property by multiplying each term of 4\sqrt{2}-2\sqrt{3} by each term of 2\sqrt{3}+5\sqrt{2}.

$$8\sqrt{3}\sqrt{2}+20\left(\sqrt{2}\right)^{2}-4\left(\sqrt{3}\right)^{2}-10\sqrt{3}\sqrt{2}$$

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

$$8\sqrt{6}+20\left(\sqrt{2}\right)^{2}-4\left(\sqrt{3}\right)^{2}-10\sqrt{3}\sqrt{2}$$

The square of \sqrt{2} is 2.

$$8\sqrt{6}+20\times 2-4\left(\sqrt{3}\right)^{2}-10\sqrt{3}\sqrt{2}$$

Multiply 20 and 2 to get 40.

$$8\sqrt{6}+40-4\left(\sqrt{3}\right)^{2}-10\sqrt{3}\sqrt{2}$$

The square of \sqrt{3} is 3.

$$8\sqrt{6}+40-4\times 3-10\sqrt{3}\sqrt{2}$$

Multiply -4 and 3 to get -12.

$$8\sqrt{6}+40-12-10\sqrt{3}\sqrt{2}$$

Subtract 12 from 40 to get 28.

$$8\sqrt{6}+28-10\sqrt{3}\sqrt{2}$$

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

$$8\sqrt{6}+28-10\sqrt{6}$$

Combine 8\sqrt{6} and -10\sqrt{6} to get -2\sqrt{6}.

$$-2\sqrt{6}+28$$

Factor

$$2 {(14 - \sqrt{6})} = 23.101020514$$