Solve (7)(-4x^2+6x-8)*(-1/2x)

$$( 7 ) ( - 4 x ^ { 2 } + 6 x - 8 ) \cdot ( - \frac { 1 } { 2 x } )$$

Evaluate

$$14x-21+\frac{28}{x}$$

Short Solution Steps

Express 7\left(-\frac{1}{2x}\right) as a single fraction.

$$\frac{-7}{2x}\left(-4x^{2}+6x-8\right)$$

Use the distributive property to multiply \frac{-7}{2x} by -4x^{2}+6x-8.

$$-4\times \frac{-7}{2x}x^{2}+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Express -4\times \frac{-7}{2x} as a single fraction.

$$\frac{-4\left(-7\right)}{2x}x^{2}+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-7\left(-1\right)\times 2}{x}x^{2}+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Express \frac{-7\left(-1\right)\times 2}{x}x^{2} as a single fraction.

$$\frac{-7\left(-1\right)\times 2x^{2}}{x}+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Cancel out x in both numerator and denominator.

$$-7\left(-1\right)\times 2x+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Express 6\times \frac{-7}{2x} as a single fraction.

$$-7\left(-1\right)\times 2x+\frac{6\left(-7\right)}{2x}x-8\times \frac{-7}{2x}$$

Cancel out 2 in both numerator and denominator.

$$-7\left(-1\right)\times 2x+\frac{-7\times 3}{x}x-8\times \frac{-7}{2x}$$

Cancel out x and x.

$$-7\left(-1\right)\times 2x-7\times 3-8\times \frac{-7}{2x}$$

Express -8\times \frac{-7}{2x} as a single fraction.

$$-7\left(-1\right)\times 2x-7\times 3+\frac{-8\left(-7\right)}{2x}$$

Cancel out 2 in both numerator and denominator.

$$-7\left(-1\right)\times 2x-7\times 3+\frac{-7\left(-1\right)\times 4}{x}$$

Do the multiplications.

$$7\times 2x-21+\frac{-7\left(-1\right)\times 4}{x}$$

Multiply 7 and 2 to get 14.

$$14x-21+\frac{-7\left(-1\right)\times 4}{x}$$

Multiply -7 and -1 to get 7.

$$14x-21+\frac{7\times 4}{x}$$

Multiply 7 and 4 to get 28.

$$14x-21+\frac{28}{x}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 14x-21 times \frac{x}{x}.

$$\frac{\left(14x-21\right)x}{x}+\frac{28}{x}$$

Since \frac{\left(14x-21\right)x}{x} and \frac{28}{x} have the same denominator, add them by adding their numerators.

$$\frac{\left(14x-21\right)x+28}{x}$$

Do the multiplications in \left(14x-21\right)x+28.

$$\frac{14x^{2}-21x+28}{x}$$

Expand

$$14x-21+\frac{28}{x}$$

Short Solution Steps

Express 7\left(-\frac{1}{2x}\right) as a single fraction.

$$\frac{-7}{2x}\left(-4x^{2}+6x-8\right)$$

Use the distributive property to multiply \frac{-7}{2x} by -4x^{2}+6x-8.

$$-4\times \frac{-7}{2x}x^{2}+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Express -4\times \frac{-7}{2x} as a single fraction.

$$\frac{-4\left(-7\right)}{2x}x^{2}+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-7\left(-1\right)\times 2}{x}x^{2}+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Express \frac{-7\left(-1\right)\times 2}{x}x^{2} as a single fraction.

$$\frac{-7\left(-1\right)\times 2x^{2}}{x}+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Cancel out x in both numerator and denominator.

$$-7\left(-1\right)\times 2x+6\times \frac{-7}{2x}x-8\times \frac{-7}{2x}$$

Express 6\times \frac{-7}{2x} as a single fraction.

$$-7\left(-1\right)\times 2x+\frac{6\left(-7\right)}{2x}x-8\times \frac{-7}{2x}$$

Cancel out 2 in both numerator and denominator.

$$-7\left(-1\right)\times 2x+\frac{-7\times 3}{x}x-8\times \frac{-7}{2x}$$

Cancel out x and x.

$$-7\left(-1\right)\times 2x-7\times 3-8\times \frac{-7}{2x}$$

Express -8\times \frac{-7}{2x} as a single fraction.

$$-7\left(-1\right)\times 2x-7\times 3+\frac{-8\left(-7\right)}{2x}$$

Cancel out 2 in both numerator and denominator.

$$-7\left(-1\right)\times 2x-7\times 3+\frac{-7\left(-1\right)\times 4}{x}$$

Do the multiplications.

$$7\times 2x-21+\frac{-7\left(-1\right)\times 4}{x}$$

Multiply 7 and 2 to get 14.

$$14x-21+\frac{-7\left(-1\right)\times 4}{x}$$

Multiply -7 and -1 to get 7.

$$14x-21+\frac{7\times 4}{x}$$

Multiply 7 and 4 to get 28.

$$14x-21+\frac{28}{x}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 14x-21 times \frac{x}{x}.

$$\frac{\left(14x-21\right)x}{x}+\frac{28}{x}$$

Since \frac{\left(14x-21\right)x}{x} and \frac{28}{x} have the same denominator, add them by adding their numerators.

$$\frac{\left(14x-21\right)x+28}{x}$$

Do the multiplications in \left(14x-21\right)x+28.

$$\frac{14x^{2}-21x+28}{x}$$