Solve operatorname{op}(-8)+op(+4)*[-3+|(-8)|]+|-6|*[operatorname{cop}80:|-20|]

$$\operatorname { op } ( - 8 ) + o p ( + 4 ) \cdot [ - 3 + | ( - 8 ) | ] + | - 6 | \cdot [ \operatorname { cop } 80 : | - 20 | ]$$

Factor

$$\frac{3o\left(40p+cp_{80}\right)}{10}$$

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -8 is 8.

$$factor(op\left(-8\right)+op\times 4\left(-3+8\right)+|-6|\times \frac{cop_{80}}{|-20|})$$

Add -3 and 8 to get 5.

$$factor(op\left(-8\right)+op\times 4\times 5+|-6|\times \frac{cop_{80}}{|-20|})$$

Multiply 4 and 5 to get 20.

$$factor(op\left(-8\right)+op\times 20+|-6|\times \frac{cop_{80}}{|-20|})$$

Combine op\left(-8\right) and op\times 20 to get 12op.

$$factor(12op+|-6|\times \frac{cop_{80}}{|-20|})$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -6 is 6.

$$factor(12op+6\times \frac{cop_{80}}{|-20|})$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -20 is 20.

$$factor(12op+6\times \frac{cop_{80}}{20})$$

Express 6\times \frac{cop_{80}}{20} as a single fraction.

$$factor(12op+\frac{6cop_{80}}{20})$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 12op times \frac{20}{20}.

$$factor(\frac{20\times 12op}{20}+\frac{6cop_{80}}{20})$$

Since \frac{20\times 12op}{20} and \frac{6cop_{80}}{20} have the same denominator, add them by adding their numerators.

$$factor(\frac{20\times 12op+6cop_{80}}{20})$$

Do the multiplications in 20\times 12op+6cop_{80}.

$$factor(\frac{240op+6cop_{80}}{20})$$

Consider 240op+6cop_{80}. Factor out 6.

$$6\left(40op+cop_{80}\right)$$

Consider 40op+cop_{80}. Factor out o.

$$o\left(40p+cp_{80}\right)$$

Rewrite the complete factored expression. Simplify.

$$\frac{3o\left(40p+cp_{80}\right)}{10}$$

Evaluate

$$\frac{3o\left(40p+cp_{80}\right)}{10}$$