Solve 3/x^2+2x-24+x/x^2+3x-18

$$\frac { 3 } { x ^ { 2 } + 2 x - 24 } + \frac { x } { x ^ { 2 } + 3 x - 18 }$$

Evaluate

$$\frac{x^{2}-x-9}{\left(x-4\right)\left(x-3\right)\left(x+6\right)}$$

Short Solution Steps

Factor x^{2}+2x-24. Factor x^{2}+3x-18.

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

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

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

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

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

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

$$\frac{3x-9+x^{2}-4x}{\left(x-4\right)\left(x-3\right)\left(x+6\right)}$$

Combine like terms in 3x-9+x^{2}-4x.

$$\frac{-x-9+x^{2}}{\left(x-4\right)\left(x-3\right)\left(x+6\right)}$$

Expand \left(x-4\right)\left(x-3\right)\left(x+6\right).

$$\frac{-x-9+x^{2}}{x^{3}-x^{2}-30x+72}$$

Differentiate w.r.t. x

$$\frac{-x^{4}+2x^{3}-4x^{2}+126x-342}{\left(\left(x-4\right)\left(x-3\right)\left(x+6\right)\right)^{2}}$$