Solve (4)(sqrt{3}-2sqrt{5})(sqrt{15}+5)-(sqrt{10}-sqrt{2})^2

$$( 4 ) ( \sqrt { 3 } - 2 \sqrt { 5 } ) ( \sqrt { 15 } + 5 ) - ( \sqrt { 10 } - \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$-20\sqrt{3}-24\sqrt{5}-12\approx -100.306647611$$

Solution Steps

Use the distributive property to multiply 4 by \sqrt{3}-2\sqrt{5}.

$$\left(4\sqrt{3}-8\sqrt{5}\right)\left(\sqrt{15}+5\right)-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Use the distributive property to multiply 4\sqrt{3}-8\sqrt{5} by \sqrt{15}+5.

$$4\sqrt{3}\sqrt{15}+20\sqrt{3}-8\sqrt{5}\sqrt{15}-40\sqrt{5}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Factor 15=3\times 5. Rewrite the square root of the product \sqrt{3\times 5} as the product of square roots \sqrt{3}\sqrt{5}.

$$4\sqrt{3}\sqrt{3}\sqrt{5}+20\sqrt{3}-8\sqrt{5}\sqrt{15}-40\sqrt{5}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$4\times 3\sqrt{5}+20\sqrt{3}-8\sqrt{5}\sqrt{15}-40\sqrt{5}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Multiply 4 and 3 to get 12.

$$12\sqrt{5}+20\sqrt{3}-8\sqrt{5}\sqrt{15}-40\sqrt{5}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$12\sqrt{5}+20\sqrt{3}-8\sqrt{5}\sqrt{5}\sqrt{3}-40\sqrt{5}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Multiply \sqrt{5} and \sqrt{5} to get 5.

$$12\sqrt{5}+20\sqrt{3}-8\times 5\sqrt{3}-40\sqrt{5}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Multiply -8 and 5 to get -40.

$$12\sqrt{5}+20\sqrt{3}-40\sqrt{3}-40\sqrt{5}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Combine 20\sqrt{3} and -40\sqrt{3} to get -20\sqrt{3}.

$$12\sqrt{5}-20\sqrt{3}-40\sqrt{5}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Combine 12\sqrt{5} and -40\sqrt{5} to get -28\sqrt{5}.

$$-28\sqrt{5}-20\sqrt{3}-\left(\sqrt{10}-\sqrt{2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{10}-\sqrt{2}\right)^{2}.

$$-28\sqrt{5}-20\sqrt{3}-\left(\left(\sqrt{10}\right)^{2}-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{10} is 10.

$$-28\sqrt{5}-20\sqrt{3}-\left(10-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

$$-28\sqrt{5}-20\sqrt{3}-\left(10-2\sqrt{2}\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

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

$$-28\sqrt{5}-20\sqrt{3}-\left(10-2\times 2\sqrt{5}+\left(\sqrt{2}\right)^{2}\right)$$

Multiply -2 and 2 to get -4.

$$-28\sqrt{5}-20\sqrt{3}-\left(10-4\sqrt{5}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$-28\sqrt{5}-20\sqrt{3}-\left(10-4\sqrt{5}+2\right)$$

Add 10 and 2 to get 12.

$$-28\sqrt{5}-20\sqrt{3}-\left(12-4\sqrt{5}\right)$$

To find the opposite of 12-4\sqrt{5}, find the opposite of each term.

$$-28\sqrt{5}-20\sqrt{3}-12+4\sqrt{5}$$

Combine -28\sqrt{5} and 4\sqrt{5} to get -24\sqrt{5}.

$$-24\sqrt{5}-20\sqrt{3}-12$$