How to reduce this to have only one x ...

Question

Let $f(x)=\frac{5}{x+4} $

Reduce the difference quotient in the alternate definition of the derivative below so that you only have one x: $$\frac{f(x) - f(2)}{x-2}$$

I've gotten down to $\frac{5(10-x)}{x-2}$ . but I can't figure out how to reduce it to one $x$. Am I doing something wrong? I have tried long division, which ended up being incorrect, so I have no idea.

Answer 1

$$5(10-x)=5(2-x+8)=-5(x-2)+40$$

thus

$$\frac{5(10-x)}{x-2}=-5+\frac{40}{x-2}$$

Answer 2

Multiply up and down by $x + 4$ $$\frac{f(x) - f(2)}{x-2}=\frac{5-\frac{5}{6}(x+4)}{(x-2)(x+4)}$$ Multiply up and down by $6$ $$\frac{30-5(x+4)}{6(x-2)(x+4)} = \frac{-5x +10}{6(x-2)(x+4)} = \frac{-5(x-2)}{6(x-2)(x+4)} = -\frac{5}{6}\frac{1}{x+4}$$

Answer 3

$$\frac{f(x) - f(2)}{x-2}$$

$$f(2)=\frac{5}{6}$$

$$\frac{\frac{5}{x+4} - \frac{5}{6}}{x-2}$$ $$\frac{30-5x-20}{6(x+4)(x-2)}$$ $$\frac{5(2-x)}{6(x+4)(x-2)}$$

$$\frac{-5}{6(x+4)}$$