Solve 16div2^{3}+3^{0}*{12-4*[7-28div(3^2-5)*2]-(3^2-2^2)^2}-17

$$16 \div 2 ^ { 3 } + 3 ^ { 0 } \cdot \{ 12 - 4 \cdot [ 7 - 28 \div ( 3 ^ { 2 } - 5 ) \cdot 2 ] - ( 3 ^ { 2 } - 2 ^ { 2 } ) ^ { 2 } \} - 17$$

Evaluate

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Solution Steps

Calculate 2 to the power of 3 and get 8.

$$\frac{16}{8}+3^{0}\left(12-4\left(7-\frac{28}{3^{2}-5}\times 2\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Divide 16 by 8 to get 2.

$$2+3^{0}\left(12-4\left(7-\frac{28}{3^{2}-5}\times 2\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Calculate 3 to the power of 0 and get 1.

$$2+1\left(12-4\left(7-\frac{28}{3^{2}-5}\times 2\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Calculate 3 to the power of 2 and get 9.

$$2+1\left(12-4\left(7-\frac{28}{9-5}\times 2\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Subtract 5 from 9 to get 4.

$$2+1\left(12-4\left(7-\frac{28}{4}\times 2\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Divide 28 by 4 to get 7.

$$2+1\left(12-4\left(7-7\times 2\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Multiply 7 and 2 to get 14.

$$2+1\left(12-4\left(7-14\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Subtract 14 from 7 to get -7.

$$2+1\left(12-4\left(-7\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Multiply 4 and -7 to get -28.

$$2+1\left(12-\left(-28\right)-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

The opposite of -28 is 28.

$$2+1\left(12+28-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Add 12 and 28 to get 40.

$$2+1\left(40-\left(3^{2}-2^{2}\right)^{2}\right)-17$$

Calculate 3 to the power of 2 and get 9.

$$2+1\left(40-\left(9-2^{2}\right)^{2}\right)-17$$

Calculate 2 to the power of 2 and get 4.

$$2+1\left(40-\left(9-4\right)^{2}\right)-17$$

Subtract 4 from 9 to get 5.

$$2+1\left(40-5^{2}\right)-17$$

Calculate 5 to the power of 2 and get 25.

$$2+1\left(40-25\right)-17$$

Subtract 25 from 40 to get 15.

$$2+1\times 15-17$$

Multiply 1 and 15 to get 15.

$$2+15-17$$

Add 2 and 15 to get 17.

$$17-17$$

Subtract 17 from 17 to get 0.

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Factor

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