Solve -4^2+(3-5)*(-2)^3+3^2-(-2)^4=

$$- 4 ^ { 2 } + ( 3 - 5 ) \cdot ( - 2 ) ^ { 3 } + 3 ^ { 2 } - ( - 2 ) ^ { 4 } =$$

Evaluate

$$-7$$

Solution Steps

Calculate 4 to the power of 2 and get 16.

$$-16+\left(3-5\right)\left(-2\right)^{3}+3^{2}-\left(-2\right)^{4}$$

Subtract 5 from 3 to get -2.

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

Calculate -2 to the power of 3 and get -8.

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

Multiply -2 and -8 to get 16.

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

Add -16 and 16 to get 0.

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

Calculate 3 to the power of 2 and get 9.

$$9-\left(-2\right)^{4}$$

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

$$9-16$$

Subtract 16 from 9 to get -7.

$$-7$$

Factor

$$-7$$