Solve (-sqrt{6}-3)^2+6*(-sqrt{6}-3)+5

$$( - \sqrt { 6 } - 3 ) ^ { 2 } + 6 \times ( - \sqrt { 6 } - 3 ) + 5$$

Evaluate

$$2$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-\sqrt{6}-3\right)^{2}.

$$\left(-\sqrt{6}\right)^{2}-6\left(-\sqrt{6}\right)+9+6\left(-\sqrt{6}-3\right)+5$$

Calculate -\sqrt{6} to the power of 2 and get \left(\sqrt{6}\right)^{2}.

$$\left(\sqrt{6}\right)^{2}-6\left(-\sqrt{6}\right)+9+6\left(-\sqrt{6}-3\right)+5$$

Multiply -6 and -1 to get 6.

$$\left(\sqrt{6}\right)^{2}+6\sqrt{6}+9+6\left(-\sqrt{6}-3\right)+5$$

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

$$\left(\sqrt{6}\right)^{2}+6\sqrt{6}+9+6\left(-\sqrt{6}\right)-18+5$$

Subtract 18 from 9 to get -9.

$$\left(\sqrt{6}\right)^{2}+6\sqrt{6}-9+6\left(-\sqrt{6}\right)+5$$

Add -9 and 5 to get -4.

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

The square of \sqrt{6} is 6.

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

Subtract 4 from 6 to get 2.

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

Multiply 6 and -1 to get -6.

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

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

$$2$$

Factor

$$2$$