Solve sqrt{3}*sqrt{12}+(-2)^4-8^2:(-4)*(-3)=

$$\sqrt { 3 } \cdot \sqrt { 12 } + ( - 2 ) ^ { 4 } - 8 ^ { 2 } : ( - 4 ) \cdot ( - 3 ) =$$

Evaluate

$$-26$$

Solution Steps

Factor 12=3\times 4. Rewrite the square root of the product \sqrt{3\times 4} as the product of square roots \sqrt{3}\sqrt{4}.

$$\sqrt{3}\sqrt{3}\sqrt{4}+\left(-2\right)^{4}-\frac{8^{2}}{-4}\left(-3\right)$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3\sqrt{4}+\left(-2\right)^{4}-\frac{8^{2}}{-4}\left(-3\right)$$

Calculate the square root of 4 and get 2.

$$3\times 2+\left(-2\right)^{4}-\frac{8^{2}}{-4}\left(-3\right)$$

Multiply 3 and 2 to get 6.

$$6+\left(-2\right)^{4}-\frac{8^{2}}{-4}\left(-3\right)$$

Calculate -2 to the power of 4 and get 16.

$$6+16-\frac{8^{2}}{-4}\left(-3\right)$$

Add 6 and 16 to get 22.

$$22-\frac{8^{2}}{-4}\left(-3\right)$$

Calculate 8 to the power of 2 and get 64.

$$22-\frac{64}{-4}\left(-3\right)$$

Divide 64 by -4 to get -16.

$$22-\left(-16\left(-3\right)\right)$$

Multiply -16 and -3 to get 48.

$$22-48$$

Subtract 48 from 22 to get -26.

$$-26$$

Factor

$$-26$$