Given that $f(x) = 1 - 3/(x+2) + 3/(x+2)$ while $x ≠- 2$, show that $f(x) = (x^2 + x + 1)/(x+2)^2$

Question

What type of problem is this and what are the first step(s) needed to tackle it?

---

Given that: $$f(x) = 1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}, \quad \text{with }x ≠ -2$$ Show that: $$f(x) = \frac{x^2 + x +1}{(x + 2)^2}$$

It's fairly basic, clearly, but from my attempts so far it appears that finding common denominators doesn't take you from the first form to the second.

Should something be factorised?

Answer 1

No need of factorization, just evaluate the sum $$1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}=\frac{(x +2)^2-3(x +2)+3}{(x +2)^2}.$$ What do you obtain after expanding the numerator?

Answer 2

$$f(x) = 1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}; x ≠ 2$$

$$\implies f(x) = \frac{(x+2)^2}{(x+2)^2}-\frac{3(x+2)}{(x + 2)^2}+\frac{3}{(x +2)^2}$$

$$\implies f(x) = \frac{(x+2)^2-3(x+2)+3}{(x+2)^2}$$

Just expand and then add up the values in the numerator...