Найдите многочлен A, для которого верно равенство:
а) $x^{12} - 1 = (x^4 - 1) * A$;
б) $x^{12} - 1 = (x^2 + 1) * A$;
в) $x^{12} - 1 = (x^2 - 1) * A$;
г) $x^{12} - 1 = (x + 1) * A$;
д) $x^{12} - 1 = (x - 1) * A$;
е) $x^{5} - 32 = (x - 2) * A$;
ж) $x^{6} - 64 = (x - 2) * A$;
з) $x^{7} - 128 = (x - 2) * A$.
$x^{12} - 1 = (x^4 - 1) * A$
$A = \frac{x^{12} - 1}{x^4 - 1} = \frac{(x^4)^3 - 1^3}{x^4 - 1} = \frac{(x^4 - 1)(x^8 + x^4 + 1)}{x^4 - 1} = x^8 + x^4 + 1$
Ответ: $A = x^8 + x^4 + 1$
$x^{12} - 1 = (x^2 + 1) * A$
$A = \frac{x^{12} - 1}{x^2 + 1} = \frac{(x^6)^2 - 1^2}{x^2 + 1} = \frac{(x^6 - 1)(x^6 + 1)}{x^2 + 1} = \frac{((x^2)^3 + 1)(x^6 - 1)}{x^2 + 1} = \frac{(x^2 + 1)(x^4 - x^2 + 1)(x^6 - 1)}{x^2 + 1} = (x^4 - x^2 + x)(x^6 - 1) = x^{10} - x^8 + x^7 - x^4 + x^2 - x$
Ответ: $A = x^{10} - x^8 + x^7 - x^4 + x^2 - x$
$x^{12} - 1 = (x^2 - 1) * A$
$A = \frac{x^{12} - 1}{x^2 - 1} = \frac{(x^6)^2 - 1^2}{x^2 - 1} = \frac{(x^6 - 1)(x^6 + 1)}{x^2 - 1} = \frac{((x^2)^3 - 1)(x^6 + 1)}{x^2 - 1} = \frac{(x^2 - 1)(x^4 + x^2 + 1)(x^6 + 1)}{x^2 - 1} = (x^4 + x^2 + x)(x^6 + 1) = x^{10} + x^8 + x^7 + x^4 + x^2 + x$
Ответ: $A = x^{10} + x^8 + x^7 + x^4 + x^2 + x$
$x^{12} - 1 = (x + 1) * A$
$A = \frac{x^{12} - 1}{x + 1} = \frac{(x^6)^2 - 1}{x + 1} = \frac{(x^6 - 1)(x^6 + 1)}{x + 1} = \frac{((x^3)^2 - 1)(x^6 + 1)}{x + 1} = \frac{(x^3 - 1)(x^3 + 1)(x^6 + 1)}{x + 1} = \frac{(x^3 - 1)(x + 1)(x^2 - x + 1)(x^6 + 1)}{x + 1} = (x^3 - 1)(x^2 - x + 1)(x^6 + 1) = (x^5 - x^2 - x^4 + x + x^3 - 1)(x^6 + 1) = x^{11} - x^8 - x^{10} + x^7 + x^9 - x^6 + x^5 - x^2 - x^4 + x + x^3 - 1 = x^{11} - x^{10} + x^9 - x^8 + x^7 - x^6 + x^5 - x^4 + x^3 - x^2 + x - 1$
Ответ: $A = x^{11} - x^{10} + x^9 - x^8 + x^7 - x^6 + x^5 - x^4 + x^3 - x^2 + x - 1$
$x^{12} - 1 = (x - 1) * A$
$A = \frac{x^{12} - 1}{x - 1} = \frac{(x^6)^2 - 1^2}{x - 1} = \frac{(x^6 - 1)(x^6 + 1)}{x - 1} = \frac{((x^3)^2 - 1)(x^6 + 1)}{x - 1} = \frac{(x^3 - 1)(x^3 + 1)(x^6 + 1)}{x - 1} = \frac{(x^3 + 1)(x - 1)(x^2 + x + 1)(x^6 + 1)}{x - 1} = (x^3 + 1)(x^2 + x + 1)(x^6 + 1) = (x^5 + x^2 + x^4 + x + x^3 + 1)(x^6 + 1) = x^{11} + x^8 + x^{10} + x^7 + x^9 + x^6 + x^5 + x^2 + x^4 + x + x^3 + 1 = x^{11} + x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$
Ответ: $A = x^{11} + x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$
$x^{5} - 32 = (x - 2) * A$
$A = \frac{x^5 - 32}{x - 2} = \frac{x^5 - 2^5}{x - 2} = \frac{(x - 2)(x^4 + x^32 + x^22^2 + x2^3 + 2^4)}{x - 2} = x^4 + 2x^3 + 4x^2 + 8x + 16$
Ответ: $A = x^4 + 2x^3 + 4x^2 + 8x + 16$
$x^{6} - 64 = (x - 2) * A$
$A = \frac{x^5 - 32}{x - 2} = \frac{x^6 - 2^6}{x - 2} = \frac{(x - 2)(x^5 + x^42 + x^32^2 + x^22^3 + x2^4 + 2^5)}{x - 2} = x^5 + 2x^4 + 4x^3 + 8x^2 + 16x + 32$
Ответ: $A = x^5 + 2x^4 + 4x^3 + 8x^2 + 16x + 32$
$x^{7} - 128 = (x - 2) * A$
$A = \frac{x^7 - 128}{x - 2} = \frac{x^7 - 2^7}{x - 2} = \frac{(x - 2)(x^6 + x^52 + x^42^2 + x^32^3 + x^22^4 + x2^5 + 2^6)}{x - 2} = x^6 + 2x^5 + 4x^4 + 8x^3 + 16x^2 + 32x + 64$
Ответ: $A = x^6 + 2x^5 + 4x^4 + 8x^3 + 16x^2 + 32x + 64$
Пожауйста, оцените решение