第1章 1.8 | 高専の数学解説

(1)

$3x^2 - 5x + 2 = (3x-2)(x-1)$ より

$(3x^2 - 5x + 2) \div (x-1) = \frac{(3x-2)(x-1)}{x-1} = 3x-2$

$2x^3 - 7x^2 + 12x - 9 = (2x-3)(x^2-2x+3)$ より

$(2x^3 - 7x^2 + 12x - 9) \div (2x-3) = \frac{(2x-3)(x^2-2x+3)}{2x-3} = x^2-2x+3$

$4x^3 - 5x^2 + x - 66 = (x-3)(4x^2+7x+22)$ より

$(4x^3 - 5x^2 + x - 66) \div (x-3) = \frac{(x-3)(4x^2+7x+22)}{x-3} = 4x^2+7x+22$

\begin{array}{r} 2x^{2}+x+2 \\ 3x^{2}-x-1{\big)}\ \overline{6x^{4}+x^{3}+3x^{2}-x+3} \\ \underline{6x^{4}-2x^{3}-2x^{2}} \\ 3x^{3}+5x^{2}-x \\ \underline{3x^{3}-x^{2}-x} \\ 6x^{2}+3 \\ \underline{6x^{2}-2x-2} \\ 2x+5 \end{array}

これから

$(6x^4+x^3+3x^2-x+3) \div (3x^2-x-1) = 2x^2+x+2 \quad \text{余り} \quad 2x+5$