Задание 625

Решение:

\(\largeа)\) Разделим многочлен \(x^3-4x^2+x+6\) на \(x+1\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^3-4x^2+x+6\\x^3+x^2\\\hline\end{array}\begin{array}{|l}x+1\\\hline{x}^2-5x+6\end{array}\\\end{array}\\\phantom{000}\begin{array}{r}-\begin{array}{r}{-}5x^2+x\phantom{0}\\{-}5x^2-5x\\\hline\end{array}\end{array}\\\phantom{0000000000}\begin{array}{r}-\begin{array}{r}6x+6\\6x+6\\\hline\end{array}\end{array}\\\phantom{0000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Итак, \(x^3-4x^2+x+6=(x+1)(x^2-5x+6)\)

Разделим многочлен \(x^3-4x^2+x+6\) на \(x-2\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^3-4x^2+x+6\\x^3-2x^2\\\hline\end{array}\begin{array}{|l}x-2\\\hline{x}^2-2x-3\end{array}\\\end{array}\\\phantom{000}\begin{array}{r}-\begin{array}{r}{-}2x^2+x\phantom{0}\\{-}2x^2+4x\\\hline\end{array}\end{array}\\\phantom{\ 00000000}\begin{array}{r}-\begin{array}{r}{-}3x+6\\{-}3x+6\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Итак, \(x^3-4x^2+x+6=(x-2)(x^2-2x-3)\)

Разделим многочлен \(x^3-4x^2+x+6\) на \(x-3\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^3-4x^2+x+6\\x^3-3x^2\\\hline\end{array}\begin{array}{|l}x-3\\\hline{x}^2-x-2\end{array}\\\end{array}\\\phantom{0000}\begin{array}{r}-\begin{array}{r}{-}x^2+x\phantom{0}\\{-}x^2+3x\\\hline\end{array}\end{array}\\\phantom{\ 00000000}\begin{array}{r}-\begin{array}{r}{-}2x+6\\{-}2x+6\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Итак, \(x^3-4x^2+x+6=(x-3)(x^2-x-2)\)

\(\largeб)\) Разделим многочлен \(x^4+2x^3+x^2+6\) на \(x^2+x+1\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^4+2x^3+\phantom{0}x^2+0x+6\\x^4+\phantom{0}x^3+\phantom{0}x^2\\\hline\end{array}\begin{array}{|l}x^2+x+1\\\hline{x}^2+x-1\end{array}\\\end{array}\\\phantom{\ 00000}\begin{array}{r}-\begin{array}{r}x^3+0x^2+0x\\x^3+\phantom{0}x^2+\phantom{0}x\\\hline\end{array}\end{array}\\\phantom{0000000000}\begin{array}{r}-\begin{array}{r}{-}x^2-x+6\\{-}x^2-x-1\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000}\begin{array}{r}7\end{array}\end{array}\)

Итак, \(x^4+2x^3+x^2+6=(x^2+x+1)(x^2+x-1)+7\)

Разделим многочлен \(x^4+2x^3+x^2+6\) на \(x^2+x-1\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^4+2x^3+\phantom{0}x^2+0x+6\\x^4+\phantom{0}x^3-\phantom{0}x^2\\\hline\end{array}\begin{array}{|l}x^2+x-1\\\hline{x}^2+x+1\end{array}\\\end{array}\\\phantom{\ 00000}\begin{array}{r}-\begin{array}{r}x^3+2x^2+0x\\x^3+\phantom{0}x^2-\phantom{0}x\\\hline\end{array}\end{array}\\\phantom{0000000000}\begin{array}{r}-\begin{array}{r}\phantom{0}x^2+\phantom{0}x+6\\\phantom{0}x^2+\phantom{0}x-1\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000}\begin{array}{r}7\end{array}\end{array}\)

Итак, \(x^4+2x^3+x^2+6=(x^2+x-1)(x^2+x+1)+7\)

Разделим многочлен \(x^4+2x^3+x^2+6\) на \(x+2\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^4+2x^3+x^2+0x+6\\x^4+2x^3\\\hline\end{array}\begin{array}{|l}x+2\\\hline{x}^3+x-2\end{array}\\\end{array}\\\phantom{\ 000000000}\begin{array}{r}-\begin{array}{r}x^2+0x\\x^2+2x\\\hline\end{array}\end{array}\\\phantom{\ 000000000000}\begin{array}{r}-\begin{array}{r}{-}2x+6\\{-}2x-4\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000000}\begin{array}{r}10\end{array}\end{array}\)

Итак, \(x^4+2x^3+x^2+6=(x+2)(x^3+x-2)+10\)

\(\largeв)\) Разделим многочлен \(x^5-1\) на \(x^4+1\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^5+0x^4+0x^3+0x^2+0x-1\\x^5+0x^4+0x^3+0x^2+\phantom{0}x\\\hline\end{array}\begin{array}{|l}x^4+0x^3+0x^2+0x+1\\\hline{x}\end{array}\\\end{array}\\\phantom{\ 000000000000000000000}\begin{array}{r}{-}x-1\end{array}\end{array}\)

Итак, \(x^5-1=(x^4+1)\cdot{x}-x-1\)

Разделим многочлен \(x^5-1\) на \(x^3-1\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^5+0x^4+0x^3+0x^2+0x-1\\x^5+0x^4+0x^3-\phantom{0}x^2\\\hline\end{array}\begin{array}{|l}x^3+0x^2+0x-1\\\hline{x^2}\end{array}\\\end{array}\\\phantom{\ 00000000000000000}\begin{array}{r}x^2+0x-1\end{array}\end{array}\)

Итак, \(x^5-1=(x^3-1)\cdot{x^2}+x^2-1\)

Разделим многочлен \(x^5-1\) на \(x^4+x^3+x^2+x+1\):

• \(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^5+0x^4+0x^3+0x^2+0x-1\\x^5+\phantom{0}x^4+\phantom{0}x^3+\phantom{0}x^2+\phantom{0}x\\\hline\end{array}\begin{array}{|l}x^4+x^3+x^2+x+1\\\hline{x}-1\end{array}\\\end{array}\\\phantom{0000}\begin{array}{r}-\begin{array}{r}{-}x^4-\phantom{0}x^3-\phantom{0}x^2-\phantom{0}x-1\\{-}x^4-\phantom{0}x^3-\phantom{0}x^2-\phantom{0}x-1\\\hline\end{array}\end{array}\\\phantom{000000000000000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Итак, \(x^5-1=(x^4+x^3+x^2+x+1)(x-1)\)