Задание 629

Найдите многочлен \(A,\) для которого верно равенство:

\({\largeа)}\ x^{12}-1=(x^4-1)\cdot{A};\)
\({\largeб)}\ x^{12}-1=(x^2+1)\cdot{A};\)

\({\largeв)}\ x^{12}-1=(x^2-1)\cdot{A};\)
\({\largeг)}\ x^{12}-1=(x+1)\cdot{A};\)

\({\largeд)}\ x^{12}-1=(x-1)\cdot{A};\)
\({\largeе)}\ x^5-32=(x-2)\cdot{A};\)

\({\largeж)}\ x^6-64=(x-2)\cdot{A};\)
\({\largeз)}\ x^7-128=(x-2)\cdot{A}.\)

Источник заимствования: Алгебра. 7 класс. Учебник для общеобразовательных организаций / С.М. Никольский, М.К. Потапов, Н.Н. Решетников, А.В. Шевкин – Просвещение, 2013. – 167 c. ISBN 978-5-09-027739-6

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Решение:

\({\largeа)}\ x^{12}-1=(x^4-1)\cdot{A}\)

Разделим многочлен \(x^{12}-1\) на \(x^4-1\):

\(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^{12}+0x^8+0x^4-1\\x^{12}-\phantom{0}x^8\\\hline\end{array}\begin{array}{|l}x^4-1\\\hline{x}^8+x^4+1\end{array}\\\end{array}\\\phantom{000000}\begin{array}{r}-\begin{array}{r}x^8+0x^4\\x^8-\phantom{0}x^4\\\hline\end{array}\end{array}\\\phantom{\ 00000000000}\begin{array}{r}-\begin{array}{r}x^4-1\\x^4-1\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Так как \(x^{12}-1=(x^4-1)(x^8+x^4+1),\) то \(A=x^8+x^4+1.\)

\({\largeб)}\ x^{12}-1=(x^2+1)\cdot{A}\)

Разделим многочлен \(x^{12}-1\) на \(x^2+1\):

\(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^{12}+0x^{10}+0x^8+0x^6+0x^4+0x^2-1\\x^{12}+\phantom{0}x^{10}\\\hline\end{array}\begin{array}{|l}x^2+1\\\hline{x}^{10}-x^8+x^6-x^4+x^2-1\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}{-}x^{10}+0x^8\\{-}x^{10}-\phantom{0}x^8\\\hline\end{array}\end{array}\\\phantom{000000000000}\begin{array}{r}-\begin{array}{r}x^8+0x^6\\x^8+\phantom{0}x^6\\\hline\end{array}\end{array}\\\phantom{0000000000000000}\begin{array}{r}-\begin{array}{r}{-}x^6+0x^4\\{-}x^6-\phantom{0}x^4\\\hline\end{array}\end{array}\\\phantom{00000000000000000000000}\begin{array}{r}-\begin{array}{r}x^4+0x^2\\x^4+\phantom{0}x^2\\\hline\end{array}\end{array}\\\phantom{000000000000000000000000000}\begin{array}{r}-\begin{array}{r}{-}x^2-1\\{-}x^2-1\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000000000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Так как \(x^{12}-1=(x^2+1)(x^{10}-x^8+x^6-x^4+x^2-1),\) то \(A=x^{10}-x^8+x^6-x^4+x^2-1.\)

\({\largeв)}\ x^{12}-1=(x^2-1)\cdot{A}\)

Разделим многочлен \(x^{12}-1\) на \(x^2-1\):

\(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^{12}+0x^{10}+0x^8+0x^6+0x^4+0x^2-1\\x^{12}-\phantom{0}x^{10}\\\hline\end{array}\begin{array}{|l}x^2-1\\\hline{x}^{10}+x^8+x^6+x^4+x^2+1\end{array}\\\end{array}\\\phantom{000000}\begin{array}{r}-\begin{array}{r}x^{10}+0x^8\\x^{10}-\phantom{0}x^8\\\hline\end{array}\end{array}\\\phantom{000000000000}\begin{array}{r}-\begin{array}{r}x^8+0x^6\\x^8-\phantom{0}x^6\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000}\begin{array}{r}-\begin{array}{r}x^6+0x^4\\x^6-\phantom{0}x^4\\\hline\end{array}\end{array}\\\phantom{00000000000000000000000}\begin{array}{r}-\begin{array}{r}x^4+0x^2\\x^4-\phantom{0}x^2\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^2-1\\x^2-1\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000000000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Так как \(x^{12}-1=(x^2-1)(x^{10}+x^8+x^6+x^4+x^2+1),\) то \(A=x^{10}+x^8+x^6+x^4+x^2+1.\)

\({\largeг)}\ x^{12}-1=(x+1)\cdot{A}\)

Разделим многочлен \(x^{12}-1\) на \(x+1\):

\(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^{12}+0x^{11}+0x^{10}+0x^9+0x^8+0x^7+0x^6+0x^5+0x^4+0x^3+0x^2+0x-1\\x^{12}+\phantom{0}x^{11}\\\hline\end{array}\begin{array}{|l}x+1\\\hline{x}^{11}-x^{10}+x^9-x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}{-}x^{11}+0x^{10}\\{-}x^{11}-\phantom{0}x^{10}\\\hline\end{array}\end{array}\\\phantom{000000000000}\begin{array}{r}-\begin{array}{r}x^{10}+0x^9\\x^{10}+\phantom{0}x^9\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000}\begin{array}{r}-\begin{array}{r}{-}x^9+0x^8\\{-}x^9-\phantom{0}x^8\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000000000}\begin{array}{r}-\begin{array}{r}x^8+0x^7\\x^8+\phantom{0}x^7\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000000000}\begin{array}{r}-\begin{array}{r}{-}x^7+0x^6\\{-}x^7-\phantom{0}x^6\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^6+0x^5\\x^6+\phantom{0}x^5\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}{-}x^5+0x^4\\{-}x^5-\phantom{0}x^4\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^4+0x^3\\x^4+\phantom{0}x^3\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}{-}x^3+0x^2\\{-}x^3-\phantom{0}x^2\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^2+0x\\x^2+\phantom{0}x\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}{-}x-1\\{-}x-1\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000000000000000000000000000000000000000000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Так как \(x^{12}-1=(x+1)(x^{11}-x^{10}+x^9-x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1),\) то \(A=x^{11}-x^{10}+x^9-x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1.\)

\({\largeд)}\ x^{12}-1=(x-1)\cdot{A}\)

Разделим многочлен \(x^{12}-1\) на \(x-1\):

\(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^{12}+0x^{11}+0x^{10}+0x^9+0x^8+0x^7+0x^6+0x^5+0x^4+0x^3+0x^2+0x-1\\x^{12}-\phantom{0}x^{11}\\\hline\end{array}\begin{array}{|l}x-1\\\hline{x}^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1\end{array}\\\end{array}\\\phantom{000000}\begin{array}{r}-\begin{array}{r}x^{11}+0x^{10}\\x^{11}-\phantom{0}x^{10}\\\hline\end{array}\end{array}\\\phantom{\ 000000000000}\begin{array}{r}-\begin{array}{r}x^{10}+0x^9\\x^{10}-\phantom{0}x^9\\\hline\end{array}\end{array}\\\phantom{0000000000000000000}\begin{array}{r}-\begin{array}{r}x^9+0x^8\\x^9-\phantom{0}x^8\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^8+0x^7\\x^8-\phantom{0}x^7\\\hline\end{array}\end{array}\\\phantom{000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^7+0x^6\\x^7-\phantom{0}x^6\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^6+0x^5\\x^6-\phantom{0}x^5\\\hline\end{array}\end{array}\\\phantom{00000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^5+0x^4\\x^5-\phantom{0}x^4\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^4+0x^3\\x^4-\phantom{0}x^3\\\hline\end{array}\end{array}\\\phantom{0000000000000000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^3+0x^2\\x^3-\phantom{0}x^2\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x^2+0x\\x^2-\phantom{0}x\\\hline\end{array}\end{array}\\\phantom{000000000000000000000000000000000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}x-1\\x-1\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000000000000000000000000000000000000000000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Так как \(x^{12}-1=(x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1),\) то \(A=x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1.\)

\({\largeе)}\ x^5-32=(x-2)\cdot{A}\)

Разделим многочлен \(x^5-32\) на \(x-2\):

\(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^5+0x^4+0x^3+0x^2+\phantom{0}0x-32\\x^5-2x^4\\\hline\end{array}\begin{array}{|l}x-2\\\hline{x}^4+2x^3+4x^2+8x+16\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}2x^4+0x^3\\2x^4-4x^3\\\hline\end{array}\end{array}\\\phantom{0000000000}\begin{array}{r}-\begin{array}{r}4x^3+0x^2\\4x^3-8x^2\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000}\begin{array}{r}-\begin{array}{l}8x^2+\phantom{0}0x\\8x^2-16x\\\hline\end{array}\end{array}\\\phantom{000000000000000000000}\begin{array}{r}-\begin{array}{r}16x-32\\16x-32\\\hline\end{array}\end{array}\\\phantom{\ 00000000000000000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Так как \(x^5-32=(x-2)(x^4+2x^3+4x^2+8x+16),\) то \(A=x^4+2x^3+4x^2+8x+16.\)

\({\largeж)}\ x^6-64=(x-2)\cdot{A}\)

Разделим многочлен \(x^6-64\) на \(x-2\):

\(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^6+0x^5+0x^4+0x^3+\phantom{0}0x^2+\phantom{0}0x-64\\x^6-2x^5\\\hline\end{array}\begin{array}{|l}x-2\\\hline{x}^5+2x^4+4x^3+8x^2+16x+32\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}2x^5+0x^4\\2x^5-4x^4\\\hline\end{array}\end{array}\\\phantom{0000000000}\begin{array}{r}-\begin{array}{r}4x^4+0x^3\\4x^4-8x^3\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000}\begin{array}{r}-\begin{array}{l}8x^3+\phantom{0}0x^2\\8x^3-16x^2\\\hline\end{array}\end{array}\\\phantom{000000000000000000000}\begin{array}{r}-\begin{array}{l}16x^2+\phantom{0}0x\\16x^2-32x\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000000000}\begin{array}{r}-\begin{array}{r}32x-64\\32x-64\\\hline\end{array}\end{array}\\\phantom{000000000000000000000000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Так как \(x^6-64=(x-2)(x^5+2x^4+4x^3+8x^2+16x+32),\) то \(A=x^5+2x^4+4x^3+8x^2+16x+32.\)

\({\largeз)}\ x^7-128=(x-2)\cdot{A}\)

Разделим многочлен \(x^7-128\) на \(x-2\):

\(\begin{array}{l}\begin{array}{r}-\begin{array}{l}x^7+0x^6+0x^5+0x^4+\phantom{0}0x^3+\phantom{0}0x^2+\phantom{0}0x-128\\x^7-2x^6\\\hline\end{array}\begin{array}{|l}x-2\\\hline{x}^6+2x^5+4x^4+8x^3+16x^2+32x+64\end{array}\\\end{array}\\\phantom{\ 0000}\begin{array}{r}-\begin{array}{r}2x^6+0x^5\\2x^6-4x^5\\\hline\end{array}\end{array}\\\phantom{0000000000}\begin{array}{r}-\begin{array}{r}4x^5+0x^4\\4x^5-8x^4\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000}\begin{array}{r}-\begin{array}{l}8x^4+\phantom{0}0x^3\\8x^4-16x^3\\\hline\end{array}\end{array}\\\phantom{000000000000000000000}\begin{array}{r}-\begin{array}{l}16x^3+\phantom{0}0x^2\\16x^3-32x^2\\\hline\end{array}\end{array}\\\phantom{\ 000000000000000000000000000}\begin{array}{r}-\begin{array}{l}32x^2+\phantom{0}0x\\32x^2-64x\\\hline\end{array}\end{array}\\\phantom{0000000000000000000000000000000000}\begin{array}{r}-\begin{array}{r}64x-128\\64x-128\\\hline\end{array}\end{array}\\\phantom{\ 0000000000000000000000000000000000000000000}\begin{array}{r}0\end{array}\end{array}\)

Так как \(x^7-128=(x-2)(x^6+2x^5+4x^4+8x^3+16x^2+32x+64),\) то \(A=x^6+2x^5+4x^4+8x^3+16x^2+32x+64.\)

Задание 628Задание 630