Solve 5/6*(-2)*2^3-2/3*(-2^2)*2^2+1

$$\frac{ 5 }{ 6 } \times (-2) \times { 2 }^{ 3 } - \frac{ 2 }{ 3 } \times ( { -2 }^{ 2 } ) \times { 2 }^{ 2 } +1$$

Evaluate

$$-23$$

Solution Steps

Express \frac{5}{6}\left(-2\right) as a single fraction.

$$\frac{5\left(-2\right)}{6}\times 2^{3}-\frac{2}{3}\left(-2\right)^{2}\times 2^{2}+1$$

Multiply 5 and -2 to get -10.

$$\frac{-10}{6}\times 2^{3}-\frac{2}{3}\left(-2\right)^{2}\times 2^{2}+1$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

$$-\frac{5}{3}\times 2^{3}-\frac{2}{3}\left(-2\right)^{2}\times 2^{2}+1$$

Calculate 2 to the power of 3 and get 8.

$$-\frac{5}{3}\times 8-\frac{2}{3}\left(-2\right)^{2}\times 2^{2}+1$$

Express -\frac{5}{3}\times 8 as a single fraction.

$$\frac{-5\times 8}{3}-\frac{2}{3}\left(-2\right)^{2}\times 2^{2}+1$$

Multiply -5 and 8 to get -40.

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

Fraction \frac{-40}{3} can be rewritten as -\frac{40}{3} by extracting the negative sign.

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

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

$$-\frac{40}{3}-\frac{2}{3}\times 4\times 2^{2}+1$$

Express \frac{2}{3}\times 4 as a single fraction.

$$-\frac{40}{3}-\frac{2\times 4}{3}\times 2^{2}+1$$

Multiply 2 and 4 to get 8.

$$-\frac{40}{3}-\frac{8}{3}\times 2^{2}+1$$

Calculate 2 to the power of 2 and get 4.

$$-\frac{40}{3}-\frac{8}{3}\times 4+1$$

Express \frac{8}{3}\times 4 as a single fraction.

$$-\frac{40}{3}-\frac{8\times 4}{3}+1$$

Multiply 8 and 4 to get 32.

$$-\frac{40}{3}-\frac{32}{3}+1$$

Since -\frac{40}{3} and \frac{32}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-40-32}{3}+1$$

Subtract 32 from -40 to get -72.

$$\frac{-72}{3}+1$$

Divide -72 by 3 to get -24.

$$-24+1$$

Add -24 and 1 to get -23.

$$-23$$

Factor

$$-23$$