Solve x^2-2x-24/x^2+11x+24+x^2-3x-28/x^2+13x+40

$$\frac { x ^ { 2 } - 2 x - 24 } { x ^ { 2 } + 11 x + 24 } + \frac { x ^ { 2 } - 3 x - 28 } { x ^ { 2 } + 13 x + 40 }$$

Evaluate

$$\frac{\left(x+4\right)\left(2x^{2}-5x-51\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Short Solution Steps

Factor x^{2}+11x+24. Factor x^{2}+13x+40.

$$\frac{x^{2}-2x-24}{\left(x+3\right)\left(x+8\right)}+\frac{x^{2}-3x-28}{\left(x+5\right)\left(x+8\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+3\right)\left(x+8\right) and \left(x+5\right)\left(x+8\right) is \left(x+3\right)\left(x+5\right)\left(x+8\right). Multiply \frac{x^{2}-2x-24}{\left(x+3\right)\left(x+8\right)} times \frac{x+5}{x+5}. Multiply \frac{x^{2}-3x-28}{\left(x+5\right)\left(x+8\right)} times \frac{x+3}{x+3}.

$$\frac{\left(x^{2}-2x-24\right)\left(x+5\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}+\frac{\left(x^{2}-3x-28\right)\left(x+3\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Since \frac{\left(x^{2}-2x-24\right)\left(x+5\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)} and \frac{\left(x^{2}-3x-28\right)\left(x+3\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-2x-24\right)\left(x+5\right)+\left(x^{2}-3x-28\right)\left(x+3\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Do the multiplications in \left(x^{2}-2x-24\right)\left(x+5\right)+\left(x^{2}-3x-28\right)\left(x+3\right).

$$\frac{x^{3}+5x^{2}-2x^{2}-10x-24x-120+x^{3}+3x^{2}-3x^{2}-9x-28x-84}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Combine like terms in x^{3}+5x^{2}-2x^{2}-10x-24x-120+x^{3}+3x^{2}-3x^{2}-9x-28x-84.

$$\frac{2x^{3}+3x^{2}-71x-204}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Expand \left(x+3\right)\left(x+5\right)\left(x+8\right).

$$\frac{2x^{3}+3x^{2}-71x-204}{x^{3}+16x^{2}+79x+120}$$

Expand

$$\frac{2x^{3}+3x^{2}-71x-204}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Short Solution Steps

Factor x^{2}+11x+24. Factor x^{2}+13x+40.

$$\frac{x^{2}-2x-24}{\left(x+3\right)\left(x+8\right)}+\frac{x^{2}-3x-28}{\left(x+5\right)\left(x+8\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+3\right)\left(x+8\right) and \left(x+5\right)\left(x+8\right) is \left(x+3\right)\left(x+5\right)\left(x+8\right). Multiply \frac{x^{2}-2x-24}{\left(x+3\right)\left(x+8\right)} times \frac{x+5}{x+5}. Multiply \frac{x^{2}-3x-28}{\left(x+5\right)\left(x+8\right)} times \frac{x+3}{x+3}.

$$\frac{\left(x^{2}-2x-24\right)\left(x+5\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}+\frac{\left(x^{2}-3x-28\right)\left(x+3\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Since \frac{\left(x^{2}-2x-24\right)\left(x+5\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)} and \frac{\left(x^{2}-3x-28\right)\left(x+3\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{2}-2x-24\right)\left(x+5\right)+\left(x^{2}-3x-28\right)\left(x+3\right)}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Do the multiplications in \left(x^{2}-2x-24\right)\left(x+5\right)+\left(x^{2}-3x-28\right)\left(x+3\right).

$$\frac{x^{3}+5x^{2}-2x^{2}-10x-24x-120+x^{3}+3x^{2}-3x^{2}-9x-28x-84}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Combine like terms in x^{3}+5x^{2}-2x^{2}-10x-24x-120+x^{3}+3x^{2}-3x^{2}-9x-28x-84.

$$\frac{2x^{3}+3x^{2}-71x-204}{\left(x+3\right)\left(x+5\right)\left(x+8\right)}$$

Expand \left(x+3\right)\left(x+5\right)\left(x+8\right).

$$\frac{2x^{3}+3x^{2}-71x-204}{x^{3}+16x^{2}+79x+120}$$