Solve 5/x^2+7x+6+2x/x^2-3x-4

$$\frac { 5 } { x ^ { 2 } + 7 x + 6 } + \frac { 2 x } { x ^ { 2 } - 3 x - 4 }$$

Evaluate

$$\frac{2x^{2}+17x-20}{\left(x-4\right)\left(x+1\right)\left(x+6\right)}$$

Short Solution Steps

Factor x^{2}+7x+6. Factor x^{2}-3x-4.

$$\frac{5}{\left(x+1\right)\left(x+6\right)}+\frac{2x}{\left(x-4\right)\left(x+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(x+6\right) and \left(x-4\right)\left(x+1\right) is \left(x-4\right)\left(x+1\right)\left(x+6\right). Multiply \frac{5}{\left(x+1\right)\left(x+6\right)} times \frac{x-4}{x-4}. Multiply \frac{2x}{\left(x-4\right)\left(x+1\right)} times \frac{x+6}{x+6}.

$$\frac{5\left(x-4\right)}{\left(x-4\right)\left(x+1\right)\left(x+6\right)}+\frac{2x\left(x+6\right)}{\left(x-4\right)\left(x+1\right)\left(x+6\right)}$$

Since \frac{5\left(x-4\right)}{\left(x-4\right)\left(x+1\right)\left(x+6\right)} and \frac{2x\left(x+6\right)}{\left(x-4\right)\left(x+1\right)\left(x+6\right)} have the same denominator, add them by adding their numerators.

$$\frac{5\left(x-4\right)+2x\left(x+6\right)}{\left(x-4\right)\left(x+1\right)\left(x+6\right)}$$

Do the multiplications in 5\left(x-4\right)+2x\left(x+6\right).

$$\frac{5x-20+2x^{2}+12x}{\left(x-4\right)\left(x+1\right)\left(x+6\right)}$$

Combine like terms in 5x-20+2x^{2}+12x.

$$\frac{17x-20+2x^{2}}{\left(x-4\right)\left(x+1\right)\left(x+6\right)}$$

Expand \left(x-4\right)\left(x+1\right)\left(x+6\right).

$$\frac{17x-20+2x^{2}}{x^{3}+3x^{2}-22x-24}$$

Differentiate w.r.t. x

$$\frac{-2x^{4}-34x^{3}-35x^{2}+24x-848}{\left(\left(x-4\right)\left(x+1\right)\left(x+6\right)\right)^{2}}$$