Solve 12{left(1/2right)}^3-40{left(1/2right)}^2+27(1/2)-5

$$12 { \left( \frac{ 1 }{ 2 } \right) }^{ 3 } -40 { \left( \frac{ 1 }{ 2 } \right) }^{ 2 } +27( \frac{ 1 }{ 2 } )-5$$

Evaluate

$$0$$

Solution Steps

Calculate \frac{1}{2} to the power of 3 and get \frac{1}{8}.

$$12\times \frac{1}{8}-40\times \left(\frac{1}{2}\right)^{2}+27\times \frac{1}{2}-5$$

Multiply 12 and \frac{1}{8} to get \frac{12}{8}.

$$\frac{12}{8}-40\times \left(\frac{1}{2}\right)^{2}+27\times \frac{1}{2}-5$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$\frac{3}{2}-40\times \left(\frac{1}{2}\right)^{2}+27\times \frac{1}{2}-5$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{3}{2}-40\times \frac{1}{4}+27\times \frac{1}{2}-5$$

Multiply 40 and \frac{1}{4} to get \frac{40}{4}.

$$\frac{3}{2}-\frac{40}{4}+27\times \frac{1}{2}-5$$

Divide 40 by 4 to get 10.

$$\frac{3}{2}-10+27\times \frac{1}{2}-5$$

Convert 10 to fraction \frac{20}{2}.

$$\frac{3}{2}-\frac{20}{2}+27\times \frac{1}{2}-5$$

Since \frac{3}{2} and \frac{20}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3-20}{2}+27\times \frac{1}{2}-5$$

Subtract 20 from 3 to get -17.

$$-\frac{17}{2}+27\times \frac{1}{2}-5$$

Multiply 27 and \frac{1}{2} to get \frac{27}{2}.

$$-\frac{17}{2}+\frac{27}{2}-5$$

Since -\frac{17}{2} and \frac{27}{2} have the same denominator, add them by adding their numerators.

$$\frac{-17+27}{2}-5$$

Add -17 and 27 to get 10.

$$\frac{10}{2}-5$$

Divide 10 by 2 to get 5.

$$5-5$$

Subtract 5 from 5 to get 0.

$$0$$

Factor

$$0$$