Solve {l}{24-4*2+6+2}{*(4+2^3div4-1)}

$$\left. \begin{array} { l } { 24 - 4 \times 2 + 6 + 2 } \\ { \times ( 4 + 2 ^ { 3 } \div 4 - 1 ) } \end{array} \right.$$

Evaluate

$$32$$

Solution Steps

Multiply 4 and 2 to get 8.

$$24-8+6+2\left(4+\frac{2^{3}}{4}-1\right)$$

Subtract 8 from 24 to get 16.

$$16+6+2\left(4+\frac{2^{3}}{4}-1\right)$$

Add 16 and 6 to get 22.

$$22+2\left(4+\frac{2^{3}}{4}-1\right)$$

Calculate 2 to the power of 3 and get 8.

$$22+2\left(4+\frac{8}{4}-1\right)$$

Divide 8 by 4 to get 2.

$$22+2\left(4+2-1\right)$$

Add 4 and 2 to get 6.

$$22+2\left(6-1\right)$$

Subtract 1 from 6 to get 5.

$$22+2\times 5$$

Multiply 2 and 5 to get 10.

$$22+10$$

Add 22 and 10 to get 32.

$$32$$

Factor

$$2^{5}$$