다항식 나눗셈의 몫

$ax ^{3} + b = \left( ax + b \right) Q _{1} \left( x \right) + R _{1}$ $\cdots \cdots \text{㉠}$ $ax ^{4} + b = \left( ax + b \right) Q _{2} \left( x \right) + R _{2}$ $\cdots \cdots \text{㉡}$ $\text{㉠}$, $\text{㉡}$에 $\displaystyle x = - \frac{b}{a}$를 각각 대입하면 $\displaystyle R _{1} = - \frac{b ^{3}}{a ^{2}} + b$, $\displaystyle R _{2} = \frac{b ^{4}}{a ^{3}} + b$ $R _{1} = R _{2}$이므로 $\displaystyle - \frac{b ^{3}}{a ^{2}} + b = \frac{b ^{4}}{a ^{3}} + b$ $\therefore$ $b = - a$ ($\because$ $ab \ne 0$) 그러므로 $R _{1} = R _{2} = 0$ $ax ^{3} - a = a \left( x - 1 \right) \left( x ^{2} + x + 1 \right)$이므로 $a \left( x - 1 \right) \left( x ^{2} + x + 1 \right) = a \left( x - 1 \right) Q _{1} ( x )$ $\therefore$ $Q _{1} ( x ) = x ^{2} + x + 1$ $ax ^{4} - a = a \left( x - 1 \right) \left( x + 1 \right) \left( x ^{2} + 1 \right)$이므로 $a \left( x - 1 \right) \left( x + 1 \right) \left( x ^{2} + 1 \right) = a \left( x - 1 \right) Q _{2} ( x )$ $\therefore$ $Q _{2} ( x ) = \left( x + 1 \right) \left( x ^{2} + 1 \right)$ 따라서 $Q _{1} \left( 2 \right) + Q _{2} \left( 1 \right) = 7 + 4 = 11$