JAMB 2001 · UME · Q11

Divide $a^{3x} - 26a^{2x} + 156a^x - 216$ by $a^{2x} - 24a^x + 108$ .

Worked solution (try it first)

Let $u = a^x$ .
Then $a^{2x} = u^2$ and $a^{3x} = u^3$ , so divide $u^3 - 26u^2 + 156u - 216$ by $u^2 - 24u + 108$ .
$u^3 \div u^2 = u$ .
Subtract $u(u^2 - 24u + 108) = u^3 - 24u^2 + 108u$ to leave $-2u^2 + 48u - 216$ .
$-2u^2 \div u^2 = -2$ , and $-2(u^2 - 24u + 108)$ is exactly that, so the remainder is 0.
The quotient is $u - 2 = a^x - 2$ , option C.