Solve (2-2sqrt{2})^2+8sqrt{2}+1

$$( 2 - 2 \sqrt { 2 } ) ^ { 2 } + 8 \sqrt { 2 } + 1$$

Evaluate

$$13$$

Solution Steps

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

$$4-8\sqrt{2}+4\left(\sqrt{2}\right)^{2}+8\sqrt{2}+1$$

The square of \sqrt{2} is 2.

$$4-8\sqrt{2}+4\times 2+8\sqrt{2}+1$$

Multiply 4 and 2 to get 8.

$$4-8\sqrt{2}+8+8\sqrt{2}+1$$

Add 4 and 8 to get 12.

$$12-8\sqrt{2}+8\sqrt{2}+1$$

Combine -8\sqrt{2} and 8\sqrt{2} to get 0.

$$12+1$$

Add 12 and 1 to get 13.

$$13$$

Factor

$$13$$