JAMB 2002 · UME · Q39

Solve for $x$ in the equation $x^3 - 5x^2 - x + 5 = 0$ .

Worked solution (try it first)

Group the terms in pairs: $x^2(x - 5) - 1(x - 5) = 0$ .
Take out the common bracket: $(x - 5)(x^2 - 1) = 0$ , and $x^2 - 1 = (x - 1)(x + 1)$ .
So $x = 5$ , $x = 1$ or $x = -1$ , option D.