Solve 2*{-1*[2+4*(3-2)]}*{2*(4-3)-2*3-4*5}=

$$2 \cdot \{ - 1 \cdot [ 2 + 4 \cdot ( 3 - 2 ) ] \} \cdot \{ 2 \cdot ( 4 - 3 ) - 2 \cdot 3 - 4 \cdot 5 \} =$$

Evaluate

$$288$$

Solution Steps

Multiply 2 and -1 to get -2.

$$-2\left(2+4\left(3-2\right)\right)\left(2\left(4-3\right)-2\times 3-4\times 5\right)$$

Subtract 2 from 3 to get 1.

$$-2\left(2+4\right)\left(2\left(4-3\right)-2\times 3-4\times 5\right)$$

Add 2 and 4 to get 6.

$$-2\times 6\left(2\left(4-3\right)-2\times 3-4\times 5\right)$$

Multiply -2 and 6 to get -12.

$$-12\left(2\left(4-3\right)-2\times 3-4\times 5\right)$$

Subtract 3 from 4 to get 1.

$$-12\left(2-2\times 3-4\times 5\right)$$

Multiply 2 and 3 to get 6.

$$-12\left(2-6-4\times 5\right)$$

Subtract 6 from 2 to get -4.

$$-12\left(-4-4\times 5\right)$$

Multiply 4 and 5 to get 20.

$$-12\left(-4-20\right)$$

Subtract 20 from -4 to get -24.

$$-12\left(-24\right)$$

Multiply -12 and -24 to get 288.

$$288$$

Factor

$$2^{5}\times 3^{2}$$