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

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

Evaluate

$$-28$$

Solution Steps

Use the distributive property to multiply 4 by \sqrt{2}+2\sqrt{3}.

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

Use the distributive property to multiply 4\sqrt{2}+8\sqrt{3} by \sqrt{2}-2\sqrt{3} and combine like terms.

$$4\left(\sqrt{2}\right)^{2}-16\left(\sqrt{3}\right)^{2}+\left(2\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4\times 2-16\left(\sqrt{3}\right)^{2}+\left(2\sqrt{3}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$8-16\left(\sqrt{3}\right)^{2}+\left(2\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$8-16\times 3+\left(2\sqrt{3}\right)^{2}$$

Multiply -16 and 3 to get -48.

$$8-48+\left(2\sqrt{3}\right)^{2}$$

Subtract 48 from 8 to get -40.

$$-40+\left(2\sqrt{3}\right)^{2}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$-40+2^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

$$-40+4\times 3$$

Multiply 4 and 3 to get 12.

$$-40+12$$

Add -40 and 12 to get -28.

$$-28$$

Factor

$$-28$$