Solve 7/x^2-36-9/x^2-2x-48-7x/x^2-14x+48

$$\frac { 7 } { x ^ { 2 } - 36 } - \frac { 9 } { x ^ { 2 } - 2 x - 48 } - \frac { 7 x } { x ^ { 2 } - 14 x + 48 }$$

Evaluate

$$-\frac{7x^{2}+44x+2}{\left(x-8\right)\left(x^{2}-36\right)}$$

Short Solution Steps

Factor x^{2}-36. Factor x^{2}-2x-48.

$$\frac{7}{\left(x-6\right)\left(x+6\right)}-\frac{9}{\left(x-8\right)\left(x+6\right)}-\frac{7x}{x^{2}-14x+48}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-6\right)\left(x+6\right) and \left(x-8\right)\left(x+6\right) is \left(x-8\right)\left(x-6\right)\left(x+6\right). Multiply \frac{7}{\left(x-6\right)\left(x+6\right)} times \frac{x-8}{x-8}. Multiply \frac{9}{\left(x-8\right)\left(x+6\right)} times \frac{x-6}{x-6}.

$$\frac{7\left(x-8\right)}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}-\frac{9\left(x-6\right)}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}-\frac{7x}{x^{2}-14x+48}$$

Since \frac{7\left(x-8\right)}{\left(x-8\right)\left(x-6\right)\left(x+6\right)} and \frac{9\left(x-6\right)}{\left(x-8\right)\left(x-6\right)\left(x+6\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7\left(x-8\right)-9\left(x-6\right)}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}-\frac{7x}{x^{2}-14x+48}$$

Do the multiplications in 7\left(x-8\right)-9\left(x-6\right).

$$\frac{7x-56-9x+54}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}-\frac{7x}{x^{2}-14x+48}$$

Combine like terms in 7x-56-9x+54.

$$\frac{-2x-2}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}-\frac{7x}{x^{2}-14x+48}$$

Factor x^{2}-14x+48.

$$\frac{-2x-2}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}-\frac{7x}{\left(x-8\right)\left(x-6\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-8\right)\left(x-6\right)\left(x+6\right) and \left(x-8\right)\left(x-6\right) is \left(x-8\right)\left(x-6\right)\left(x+6\right). Multiply \frac{7x}{\left(x-8\right)\left(x-6\right)} times \frac{x+6}{x+6}.

$$\frac{-2x-2}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}-\frac{7x\left(x+6\right)}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}$$

Since \frac{-2x-2}{\left(x-8\right)\left(x-6\right)\left(x+6\right)} and \frac{7x\left(x+6\right)}{\left(x-8\right)\left(x-6\right)\left(x+6\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-2x-2-7x\left(x+6\right)}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}$$

Do the multiplications in -2x-2-7x\left(x+6\right).

$$\frac{-2x-2-7x^{2}-42x}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}$$

Combine like terms in -2x-2-7x^{2}-42x.

$$\frac{-44x-2-7x^{2}}{\left(x-8\right)\left(x-6\right)\left(x+6\right)}$$

Expand \left(x-8\right)\left(x-6\right)\left(x+6\right).

$$\frac{-44x-2-7x^{2}}{x^{3}-8x^{2}-36x+288}$$

Differentiate w.r.t. x

$$\frac{7x^{4}+88x^{3}-94x^{2}-4064x-12744}{\left(\left(x-8\right)\left(x^{2}-36\right)\right)^{2}}$$