Solve 1*x+3/x^2-4/frac{x^2-x-12{x^2-8}})

$$\frac { 1 \cdot \frac { x + 3 } { x ^ { 2 } - 4 } } { \frac { x ^ { 2 } - x - 12 } { x ^ { 2 } - 8 } } )$$

Evaluate

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

Short Solution Steps

Express 1\times \frac{x+3}{x^{2}-4} as a single fraction.

$$\frac{\frac{x+3}{x^{2}-4}}{\frac{x^{2}-x-12}{x^{2}-8}}$$

Divide \frac{x+3}{x^{2}-4} by \frac{x^{2}-x-12}{x^{2}-8} by multiplying \frac{x+3}{x^{2}-4} by the reciprocal of \frac{x^{2}-x-12}{x^{2}-8}.

$$\frac{\left(x+3\right)\left(x^{2}-8\right)}{\left(x^{2}-4\right)\left(x^{2}-x-12\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(x+3\right)\left(x^{2}-8\right)}{\left(x-4\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)}$$

Cancel out x+3 in both numerator and denominator.

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

Expand the expression.

$$\frac{x^{2}-8}{x^{3}-4x^{2}-4x+16}$$

Expand

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

Short Solution Steps

Express 1\times \frac{x+3}{x^{2}-4} as a single fraction.

$$\frac{\frac{x+3}{x^{2}-4}}{\frac{x^{2}-x-12}{x^{2}-8}}$$

Divide \frac{x+3}{x^{2}-4} by \frac{x^{2}-x-12}{x^{2}-8} by multiplying \frac{x+3}{x^{2}-4} by the reciprocal of \frac{x^{2}-x-12}{x^{2}-8}.

$$\frac{\left(x+3\right)\left(x^{2}-8\right)}{\left(x^{2}-4\right)\left(x^{2}-x-12\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(x+3\right)\left(x^{2}-8\right)}{\left(x-4\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)}$$

Cancel out x+3 in both numerator and denominator.

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

Expand the expression.

$$\frac{x^{2}-8}{x^{3}-4x^{2}-4x+16}$$