Solve x-2/x^2+5x+4-8/x^2+12x+32

$$\frac { x - 2 } { x ^ { 2 } + 5 x + 4 } - \frac { 8 } { x ^ { 2 } + 12 x + 32 }$$

Evaluate

$$\frac{x-6}{\left(x+1\right)\left(x+8\right)}$$

Short Solution Steps

Factor x^{2}+5x+4. Factor x^{2}+12x+32.

$$\frac{x-2}{\left(x+1\right)\left(x+4\right)}-\frac{8}{\left(x+4\right)\left(x+8\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(x+4\right) and \left(x+4\right)\left(x+8\right) is \left(x+1\right)\left(x+4\right)\left(x+8\right). Multiply \frac{x-2}{\left(x+1\right)\left(x+4\right)} times \frac{x+8}{x+8}. Multiply \frac{8}{\left(x+4\right)\left(x+8\right)} times \frac{x+1}{x+1}.

$$\frac{\left(x-2\right)\left(x+8\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}-\frac{8\left(x+1\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Since \frac{\left(x-2\right)\left(x+8\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)} and \frac{8\left(x+1\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x-2\right)\left(x+8\right)-8\left(x+1\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Do the multiplications in \left(x-2\right)\left(x+8\right)-8\left(x+1\right).

$$\frac{x^{2}+8x-2x-16-8x-8}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Combine like terms in x^{2}+8x-2x-16-8x-8.

$$\frac{x^{2}-2x-24}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Factor the expressions that are not already factored in \frac{x^{2}-2x-24}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}.

$$\frac{\left(x-6\right)\left(x+4\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Cancel out x+4 in both numerator and denominator.

$$\frac{x-6}{\left(x+1\right)\left(x+8\right)}$$

Expand \left(x+1\right)\left(x+8\right).

$$\frac{x-6}{x^{2}+9x+8}$$

Expand

$$\frac{x-6}{\left(x+1\right)\left(x+8\right)}$$

Short Solution Steps

Factor x^{2}+5x+4. Factor x^{2}+12x+32.

$$\frac{x-2}{\left(x+1\right)\left(x+4\right)}-\frac{8}{\left(x+4\right)\left(x+8\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(x+4\right) and \left(x+4\right)\left(x+8\right) is \left(x+1\right)\left(x+4\right)\left(x+8\right). Multiply \frac{x-2}{\left(x+1\right)\left(x+4\right)} times \frac{x+8}{x+8}. Multiply \frac{8}{\left(x+4\right)\left(x+8\right)} times \frac{x+1}{x+1}.

$$\frac{\left(x-2\right)\left(x+8\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}-\frac{8\left(x+1\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Since \frac{\left(x-2\right)\left(x+8\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)} and \frac{8\left(x+1\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x-2\right)\left(x+8\right)-8\left(x+1\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Do the multiplications in \left(x-2\right)\left(x+8\right)-8\left(x+1\right).

$$\frac{x^{2}+8x-2x-16-8x-8}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Combine like terms in x^{2}+8x-2x-16-8x-8.

$$\frac{x^{2}-2x-24}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Factor the expressions that are not already factored in \frac{x^{2}-2x-24}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}.

$$\frac{\left(x-6\right)\left(x+4\right)}{\left(x+1\right)\left(x+4\right)\left(x+8\right)}$$

Cancel out x+4 in both numerator and denominator.

$$\frac{x-6}{\left(x+1\right)\left(x+8\right)}$$

Expand \left(x+1\right)\left(x+8\right).

$$\frac{x-6}{x^{2}+9x+8}$$