WAEC 2009 · Paper 2 · Q11✱

(a)(i)
The polynomial $f(x) = 2x^3 + px^2 + qx + r$ is divisible by $(2x^2 - 7x + 3)$ . It has a remainder of $-36$ when it is divided by $(x + 1)$ . Find the values of the constants $p$ , $q$ and $r$ ;
Separate values with commas, e.g. 3, −2
(a)(ii)
the zeros of $f(x)$ .
Separate values with commas, e.g. 3, −2
(b)
Find the truth set of $x^2 - 3x + 2 < 0$ .
Show the answer
$\{x : 1 < x < 2\}$

Worked solution (try it first)

(a)(i)

$2x^2 - 7x + 3 = (2x - 1)(x - 3)$ , so $f(\frac12) = 0$ and $f(3) = 0$ .
$f(\frac12) = 0$ : $\frac14 + \frac{p}{4} + \frac{q}{2} + r = 0$ , so $p + 2q + 4r = -1$ .
$f(3) = 0$ : $54 + 9p + 3q + r = 0$ , so $9p + 3q + r = -54$ .
Remainder theorem, $f(-1) = -36$ : $-2 + p - q + r = -36$ , so $p - q + r = -34$ .
Solve the three equations: $p = -11$ , $q = 17$ and $r = -6$ .
Check: $f(-1) = -2 - 11 - 17 - 6 = -36$ ✓.

(ii)

$f(x) = 2x^3 - 11x^2 + 17x - 6$
$= (2x^2 - 7x + 3)(x - 2)$
$= (2x - 1)(x - 3)(x - 2)$ .
So the zeros are $x = \frac12$ , $2$ and $3$ .

(b)

Factorise: $(x - 1)(x - 2) < 0$ .
The product is negative only when the brackets have opposite signs, which is between the roots.
The truth set is $\{x : 1 < x < 2\}$ .

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