Solve 2(sqrt{2}+1)^2-4(sqrt{2}+1)

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

Evaluate

$$2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+1\right)^{2}.

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

$$2\left(3+2\sqrt{2}\right)-4\left(\sqrt{2}+1\right)$$

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

$$6+4\sqrt{2}-4\left(\sqrt{2}+1\right)$$

Use the distributive property to multiply -4 by \sqrt{2}+1.

$$6+4\sqrt{2}-4\sqrt{2}-4$$

Combine 4\sqrt{2} and -4\sqrt{2} to get 0.

$$6-4$$

Subtract 4 from 6 to get 2.

$$2$$

Factor

$$2$$