Solve -3{left(1+sqrt{2}right)}^2+6(1+sqrt{2)}+9

$$-3 { \left(1+ \sqrt{ 2 } \right) }^{ 2 } +6(1+ \sqrt{ 2) } +9$$

Evaluate

$$6$$

Solution Steps

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

$$-3\left(1+2\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)+6\left(1+\sqrt{2}\right)+9$$

The square of \sqrt{2} is 2.

$$-3\left(1+2\sqrt{2}+2\right)+6\left(1+\sqrt{2}\right)+9$$

Add 1 and 2 to get 3.

$$-3\left(3+2\sqrt{2}\right)+6\left(1+\sqrt{2}\right)+9$$

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

$$-9-6\sqrt{2}+6\left(1+\sqrt{2}\right)+9$$

Use the distributive property to multiply 6 by 1+\sqrt{2}.

$$-9-6\sqrt{2}+6+6\sqrt{2}+9$$

Add -9 and 6 to get -3.

$$-3-6\sqrt{2}+6\sqrt{2}+9$$

Combine -6\sqrt{2} and 6\sqrt{2} to get 0.

$$-3+9$$

Add -3 and 9 to get 6.

$$6$$

Factor

$$2\times 3$$