数学 – 文字と記号の活躍 - 整式の除法と分数式 – 整式の最大公約数と最小公倍数(ユークリッドの互除法)

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数学読本〈1〉数・式の計算/方程式/不等式 (松坂和夫(著)、岩波書店)の第2章(文字と記号の活躍 - 式の計算)、2.3(整式の除法と分数式)、整式の最大公約数と最小公倍数の問18.を解いてみる。

問18.

$\begin{array}{l} x^{3} - 4 x^{2} - 16 x - 35 = (x^{3} - 5 x^{2} - 8 x - 42) + x^{2} - 8 x + 7 \\ x^{3} - 5 x^{2} - 8 x - 42 = (x^{2} - 8 x + 7) x + 3 x^{2} - 15 x - 42 \\ x^{2} - 8 x + 7 = (x^{2} - 5 x - 14) - 3 x + 21 \\ x^{2} - 5 x - 14 = (x - 7) x + 2 x - 14 \\ x - 7 = (x - 7) \cdot 1 \\ x - 7 \end{array}$
$\begin{array}{l} x^{4} + 4 x^{3} + 3 x^{2} + 4 x + 4 = (x^{3} + 3 x^{2} + 6 x + 4) x + x^{3} - 3 x^{2} + 4 \\ x^{3} + 3 x^{2} + 6 x + 4 = (x^{3} - 3 x^{2} + 4) \cdot 1 + 6 x^{2} + 6 x \\ x^{3} - 3 x^{2} + 4 = (x^{2} + x) x - 4 x^{2} + 4 \\ x^{2} + x = (x^{2} - 1) \cdot 1 + x + 1 \\ x^{2} - 1 = (x + 1) x - x - 1 \\ x + 1 = (x + 1) \cdot 1 \\ x + 1 \end{array}$
$\begin{array}{l} x^{5} - 5 x^{4} + 7 x^{3} + x^{2} - 8 x - 2 = (x^{4} - 2 x^{2} - 12 x - 8) (x - 5) + 9 x^{3} + 3 x^{2} - 60 x - 42 \\ x^{5} - 5 x^{4} + 7 x^{3} + x^{2} - 8 x - 2 = (x^{4} - 2 x^{2} - 12 x - 8) (x - 5) + 3 (3 x^{3} + x^{2} - 20 x - 14) \\ 3 x^{4} - 6 x^{2} - 36 x - 24 = (3 x^{3} + x^{2} - 20 x - 14) x - x^{3} + 14 x^{2} - 22 x - 24 \\ 3 x^{4} - 6 x^{2} - 36 x - 24 = (3 x^{3} + x^{2} - 20 x - 14) x - 3 x^{3} + 42 x^{2} - 66 x - 72 \\ 3 x^{4} - 6 x^{2} - 36 x - 24 = (3 x^{3} + x^{2} - 20 x - 14) (x - 1) + 43 x^{2} - 86 x - 86 \\ 3 x^{4} - 6 x^{2} - 36 x - 24 = (3 x^{3} + x^{2} - 20 x - 14) (x - 1) + x^{2} - 2 x - 2 \\ 3 x^{3} + x^{2} - 20 x - 14 = (x^{2} - 2 x - 2) 3 x + 7 x^{2} - 14 x - 14 \\ 3 x^{3} + x^{2} - 20 x - 14 = (x^{2} - 2 x - 2) (3 x - 7) \\ x^{2} - 2 x - 2 \end{array}$

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