WAEC 2008 · Paper 2 · Q11

Worked solution (try it first)

Factorise the denominator: $x^3 - 4x^2 + 3x = x(x^2 - 4x + 3)$
$= x(x - 1)(x - 3)$ .
Write $\dfrac{2x^2 - 5x + 1}{x(x - 1)(x - 3)} = \dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x - 3}$ .
Multiply through by the denominator: $2x^2 - 5x + 1 = A(x - 1)(x - 3) + Bx(x - 3) + Cx(x - 1)$ .
Put $x = 0$ : $1 = 3A$ , so $A = \frac13$ .
Put $x = 1$ : $2 - 5 + 1 = B(1)(-2)$ , so $-2 = -2B$ and $B = 1$ .
Put $x = 3$ : $18 - 15 + 1 = C(3)(2)$ , so $4 = 6C$ and $C = \frac23$ .
So the expression is $\dfrac{1}{3x} + \dfrac{1}{x - 1} + \dfrac{2}{3(x - 3)}$ .