최대공약수 조건

[출제의도] 다항식의 최대공약수를 활용하여 추론하기 (가), (나)에서 두 다항식 $A \left( x \right)$, $B \left( x \right)$는 $A \left( x \right) = p \left( x \right) + q \left( x \right) = 2 x \left( x - 1 \right) \left( x + a \right)$ $B \left( x \right) = p \left( x \right) - q \left( x \right) = \left( x + 1 \right) \left( bx + 4 \right)$ (다)에서 $A \left( x \right)$와 $B \left( x \right)$는 $p \left( x \right)$와 $q \left( x \right)$의 최대공약수를 인수로 갖는다. $\therefore$$A \left( x \right)$와 $B \left( x \right)$는 $x + 1$과 $x - 1$을 인수로 갖는다. 인수정리에 의해 $A \left( x \right) = p \left( x \right) + q \left( x \right) = 2 x \left( x - 1 \right) \left( x + a \right)$ $A \left( - 1 \right) = \left( - 2 \right) \times \left( - 2 \right) \times \left( - 1 + a \right) = 0$ $\therefore$ $a = 1$ $\therefore$ $A \left( x \right) = p \left( x \right) + q \left( x \right) = 2 x \left( x + 1 \right) \left( x - 1 \right)$ $B \left( x \right) = p \left( x \right) - q \left( x \right) = \left( x + 1 \right) \left( bx + 4 \right)$ $B \left( 1 \right) = \left( 1 + 1 \right) \left( b + 4 \right) = 0$ $\therefore$ $b = - 4$ $\therefore$ $B \left( x \right) = p \left( x \right) - q \left( x \right) = - 4 \left( x + 1 \right) \left( x - 1 \right)$ 따라서 $\displaystyle A \left( 2 \right) \times B \left( \frac{1}{2} \right) = 36$