Solve (sqrt{7}-3)^2+6(sqrt{7}-3)+9

$$( \sqrt { 7 } - 3 ) ^ { 2 } + 6 ( \sqrt { 7 } - 3 ) + 9$$

Evaluate

$$7$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7}-3\right)^{2}.

$$\left(\sqrt{7}\right)^{2}-6\sqrt{7}+9+6\left(\sqrt{7}-3\right)+9$$

The square of \sqrt{7} is 7.

$$7-6\sqrt{7}+9+6\left(\sqrt{7}-3\right)+9$$

Add 7 and 9 to get 16.

$$16-6\sqrt{7}+6\left(\sqrt{7}-3\right)+9$$

Use the distributive property to multiply 6 by \sqrt{7}-3.

$$16-6\sqrt{7}+6\sqrt{7}-18+9$$

Combine -6\sqrt{7} and 6\sqrt{7} to get 0.

$$16-18+9$$

Subtract 18 from 16 to get -2.

$$-2+9$$

Add -2 and 9 to get 7.

$$7$$

Factor

$$7$$