Solve 40-2[3-8+(12+3)+4-(8-6)]

$$40 - 2 [ 3 - 8 + ( 12 + 3 ) + 4 - ( 8 - 6 ) ]$$

Evaluate

$$16$$

Solution Steps

Subtract 8 from 3 to get -5.

$$40-2\left(-5+12+3+4-\left(8-6\right)\right)$$

Add -5 and 12 to get 7.

$$40-2\left(7+3+4-\left(8-6\right)\right)$$

Add 7 and 3 to get 10.

$$40-2\left(10+4-\left(8-6\right)\right)$$

Add 10 and 4 to get 14.

$$40-2\left(14-\left(8-6\right)\right)$$

Subtract 6 from 8 to get 2.

$$40-2\left(14-2\right)$$

Subtract 2 from 14 to get 12.

$$40-2\times 12$$

Multiply 2 and 12 to get 24.

$$40-24$$

Subtract 24 from 40 to get 16.

$$16$$

Factor

$$2^{4}$$