Solve x+4/x^3-8+1/2x-x^2=

$$\frac { x + 4 } { x ^ { 3 } - 8 } + \frac { 1 } { 2 x - x ^ { 2 } } =$$

Evaluate

$$\frac{2}{x\left(x^{2}+2x+4\right)}$$

Short Solution Steps

Factor x^{3}-8. Factor 2x-x^{2}.

$$\frac{x+4}{\left(x-2\right)\left(x^{2}+2x+4\right)}+\frac{1}{x\left(-x+2\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-2\right)\left(x^{2}+2x+4\right) and x\left(-x+2\right) is x\left(x-2\right)\left(x^{2}+2x+4\right). Multiply \frac{x+4}{\left(x-2\right)\left(x^{2}+2x+4\right)} times \frac{x}{x}. Multiply \frac{1}{x\left(-x+2\right)} times \frac{-\left(x^{2}+2x+4\right)}{-\left(x^{2}+2x+4\right)}.

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

Since \frac{\left(x+4\right)x}{x\left(x-2\right)\left(x^{2}+2x+4\right)} and \frac{-\left(x^{2}+2x+4\right)}{x\left(x-2\right)\left(x^{2}+2x+4\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(x+4\right)x-\left(x^{2}+2x+4\right).

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

Combine like terms in x^{2}+4x-x^{2}-2x-4.

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

Factor the expressions that are not already factored in \frac{2x-4}{x\left(x-2\right)\left(x^{2}+2x+4\right)}.

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

Cancel out x-2 in both numerator and denominator.

$$\frac{2}{x\left(x^{2}+2x+4\right)}$$

Expand x\left(x^{2}+2x+4\right).

$$\frac{2}{x^{3}+2x^{2}+4x}$$

Expand

$$\frac{2}{x\left(x^{2}+2x+4\right)}$$

Short Solution Steps

Factor x^{3}-8. Factor 2x-x^{2}.

$$\frac{x+4}{\left(x-2\right)\left(x^{2}+2x+4\right)}+\frac{1}{x\left(-x+2\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-2\right)\left(x^{2}+2x+4\right) and x\left(-x+2\right) is x\left(x-2\right)\left(x^{2}+2x+4\right). Multiply \frac{x+4}{\left(x-2\right)\left(x^{2}+2x+4\right)} times \frac{x}{x}. Multiply \frac{1}{x\left(-x+2\right)} times \frac{-\left(x^{2}+2x+4\right)}{-\left(x^{2}+2x+4\right)}.

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

Since \frac{\left(x+4\right)x}{x\left(x-2\right)\left(x^{2}+2x+4\right)} and \frac{-\left(x^{2}+2x+4\right)}{x\left(x-2\right)\left(x^{2}+2x+4\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(x+4\right)x-\left(x^{2}+2x+4\right).

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

Combine like terms in x^{2}+4x-x^{2}-2x-4.

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

Factor the expressions that are not already factored in \frac{2x-4}{x\left(x-2\right)\left(x^{2}+2x+4\right)}.

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

Cancel out x-2 in both numerator and denominator.

$$\frac{2}{x\left(x^{2}+2x+4\right)}$$

Expand x\left(x^{2}+2x+4\right).

$$\frac{2}{x^{3}+2x^{2}+4x}$$