JAMB 1994 · UME · Q12

Find the values of $p$ and $q$ such that $(x - 1)$ and $(x - 3)$ are factors of $px^3 + qx^2 + 11x - 6$ .

Worked solution (try it first)

By the factor theorem, the expression is 0 at $x = 1$ : $p + q + 11 - 6 = 0$ , so $p + q = -5$ .
It is 0 at $x = 3$ : $27p + 9q + 33 - 6 = 0$ .
Divide by 9: $3p + q = -3$ .
Subtract the first equation: $2p = 2$ , so $p = 1$ and $q = -6$ , option B.