Solve (sqrt{3}+2)^2-4(sqrt{3}+2)+1

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

Evaluate

$$0$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+2\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 4 to get 7.

$$7+4\sqrt{3}-4\left(\sqrt{3}+2\right)+1$$

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

$$7+4\sqrt{3}-4\sqrt{3}-8+1$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$7-8+1$$

Subtract 8 from 7 to get -1.

$$-1+1$$

Add -1 and 1 to get 0.

$$0$$

Factor

$$0$$