Solve -x-2/x^2-2x-8-x-1/x^2-4

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

Evaluate

$$\frac{-2x^{2}+9x-8}{\left(x-4\right)\left(x^{2}-4\right)}$$

Short Solution Steps

Factor x^{2}-2x-8. Factor x^{2}-4.

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

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

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

Since -\frac{\left(x-2\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)\left(x+2\right)} and \frac{\left(x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in -\left(x-2\right)\left(x-2\right)-\left(x-1\right)\left(x-4\right).

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

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

$$\frac{-2x^{2}+9x-8}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}$$

Expand \left(x-4\right)\left(x-2\right)\left(x+2\right).

$$\frac{-2x^{2}+9x-8}{x^{3}-4x^{2}-4x+16}$$

Expand

$$\frac{-2x^{2}+9x-8}{\left(x-4\right)\left(x^{2}-4\right)}$$

Short Solution Steps

Factor x^{2}-2x-8. Factor x^{2}-4.

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

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

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

Since -\frac{\left(x-2\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)\left(x+2\right)} and \frac{\left(x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in -\left(x-2\right)\left(x-2\right)-\left(x-1\right)\left(x-4\right).

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

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

$$\frac{-2x^{2}+9x-8}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}$$

Expand \left(x-4\right)\left(x-2\right)\left(x+2\right).

$$\frac{-2x^{2}+9x-8}{x^{3}-4x^{2}-4x+16}$$