Question

x⁴+5x³+2x²+5x+1 Найдите корни симметрического много члена​

Answer 1

$x^4+5x^3+2x^2+5x+1 =0\\x^4+5x^3+x^2+x^2+5x+1=0\\x^2(x^2+1)+5x(x^2+1)+x^2+1=0\\(x^2+1)(x^2+5x+1)=0\\\\x^2+1=0\\x \in R\\\\x^2+5x+1=0\\\\x_1 = \dfrac{-5+\sqrt{21}}{2}\\\\x_2 = \dfrac{-5-\sqrt{21}}{2}$