Solve 6/x^2-16-2/x^2-x-12

$$\frac { 6 } { x ^ { 2 } - 16 } - \frac { 2 } { x ^ { 2 } - x - 12 }$$

Evaluate

$$\frac{2\left(2x+5\right)}{\left(x+3\right)\left(x^{2}-16\right)}$$

Short Solution Steps

Factor x^{2}-16. Factor x^{2}-x-12.

$$\frac{6}{\left(x-4\right)\left(x+4\right)}-\frac{2}{\left(x-4\right)\left(x+3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-4\right)\left(x+4\right) and \left(x-4\right)\left(x+3\right) is \left(x-4\right)\left(x+3\right)\left(x+4\right). Multiply \frac{6}{\left(x-4\right)\left(x+4\right)} times \frac{x+3}{x+3}. Multiply \frac{2}{\left(x-4\right)\left(x+3\right)} times \frac{x+4}{x+4}.

$$\frac{6\left(x+3\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}-\frac{2\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}$$

Since \frac{6\left(x+3\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)} and \frac{2\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6\left(x+3\right)-2\left(x+4\right)}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}$$

Do the multiplications in 6\left(x+3\right)-2\left(x+4\right).

$$\frac{6x+18-2x-8}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}$$

Combine like terms in 6x+18-2x-8.

$$\frac{4x+10}{\left(x-4\right)\left(x+3\right)\left(x+4\right)}$$

Expand \left(x-4\right)\left(x+3\right)\left(x+4\right).

$$\frac{4x+10}{x^{3}+3x^{2}-16x-48}$$

Factor

$$\frac{2\left(2x+5\right)}{\left(x+3\right)\left(x^{2}-16\right)}$$