WAEC 2018 · Paper 2 · Q10✱✱

(a)
The function $f(x) = px^2 + qx + r$ , where $p$ , $q$ and $r$ are constants. If $f(1) = 0$ , $f(-1) = 4$ and $f(2) = 7$ , find the: (i) values of $p$ , $q$ and $r$ ; (ii) factors of $f(x)$ .
Separate values with commas, e.g. 3, −2
(b)
A function $f$ is defined by $f(x) = \dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 2)}$ . Express $f(x)$ in partial fractions.

Worked solution (try it first)

(a)(i)

Each value gives an equation: $f(1) = p + q + r = 0$ , $f(-1) = p - q + r = 4$ and $f(2) = 4p + 2q + r = 7$ .
Take the second equation from the first: $2q = -4$ , so $q = -2$ .
Put $q = -2$ in the first: $p + r = 2$ .
Put it in the third: $4p + r = 11$ .
Take $p + r = 2$ from $4p + r = 11$ : $3p = 9$ , so $p = 3$ .
Then $r = 2 - 3 = -1$ .

(ii)

$f(x) = 3x^2 - 2x - 1$ .
Two numbers that multiply to $-3$ and add to $-2$ are $-3$ and 1.
So $3x^2 - 3x + x - 1 = 3x(x - 1) + 1(x - 1)$
$= (x - 1)(3x + 1)$ .

(b)

Write $\dfrac{2x^2 - x + 3}{(x + 1)(x^2 + 2)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 2}$ , so $2x^2 - x + 3 = A(x^2 + 2) + (Bx + C)(x + 1)$ .
Put $x = -1$ : $2 + 1 + 3 = 3A$ , so $A = 2$ .
Compare the $x^2$ terms: $2 = A + B$ , so $B = 0$ .
Compare the constants: $3 = 2A + C$ , so $C = -1$ .
So $f(x) = \dfrac{2}{x + 1} - \dfrac{1}{x^2 + 2}$ .

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