Solve 2/x^2+x-12-8/x^2-9

$$\frac { 2 } { x ^ { 2 } + x - 12 } - \frac { 8 } { x ^ { 2 } - 9 }$$

Evaluate

$$-\frac{2\left(3x+13\right)}{\left(x+4\right)\left(x^{2}-9\right)}$$

Short Solution Steps

Factor x^{2}+x-12. Factor x^{2}-9.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)\left(x+4\right) and \left(x-3\right)\left(x+3\right) is \left(x-3\right)\left(x+3\right)\left(x+4\right). Multiply \frac{2}{\left(x-3\right)\left(x+4\right)} times \frac{x+3}{x+3}. Multiply \frac{8}{\left(x-3\right)\left(x+3\right)} times \frac{x+4}{x+4}.

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

Since \frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)} and \frac{8\left(x+4\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(x+3\right)-8\left(x+4\right).

$$\frac{2x+6-8x-32}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}$$

Combine like terms in 2x+6-8x-32.

$$\frac{-6x-26}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}$$

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

$$\frac{-6x-26}{x^{3}+4x^{2}-9x-36}$$

Factor

$$-\frac{2\left(3x+13\right)}{\left(x+4\right)\left(x^{2}-9\right)}$$