JAMB 1995 · UME · Q18

Simplify $\dfrac{x^2 - 1}{x^3 + 2x^2 - x - 2}$ .

Worked solution (try it first)

Group the bottom in pairs: $x^3 + 2x^2 - x - 2 = x^2(x + 2) - (x + 2)$ , which is $(x + 2)(x^2 - 1)$ .
So the fraction is $\dfrac{x^2 - 1}{(x + 2)(x^2 - 1)}$ .
Cancel $x^2 - 1$ .
This leaves $\dfrac{1}{x + 2}$ , option A.