Представьте многочлен x^4-3x^2+1 в виде произведения. Вычислите 12,5^2+1,2^2-...

Решите задачу:

$x^4-3x^2+1=x^4-x^3-x^2+x^3-x^2-x-x^2+x+1= \\ =x^2(x^2-x-1)+x(x^2-x-1)-1(x^2-x-1)= \\ =(x^2+x-1)(x^2-x-1)$

$12,5^2+1,2^2-[(12,5^3+1,2^3):13,7]= \\ \\ =(\frac{125}{10})^2+(\frac{12}{10})^2-[((\frac{125}{10})^3+(\frac{12}{10})^3):\frac{137}{10}]= \\ \\ =(\frac{25}{2})^2+(\frac{6}{5})^2-[((\frac{25}{2})^3+(\frac{6}{5})^3):\frac{137}{10}]= \\ \\ =\frac{625}{4}+\frac{36}{25}-[(\frac{15625}{8}+\frac{216}{125}):\frac{137}{10}]= \\ \\ =\frac{625*25+36*4}{100}-[\frac{15625*125+216*8}{1000}:\frac{137}{10}]= \\ \\ =\frac{15769}{100}-[\frac{1954853}{1000}:\frac{137}{10}]=$

$=\frac{15769}{100}-[\frac{1954853}{1000}*\frac{10}{137}]= \\ \\ =\frac{15769}{100}-\frac{14269}{100}= \frac{1500}{100}=15$