x^4-4x³+4x²-9=0 решите плиз

x^4 - 4x^3 + 4x^2 - 9 = 0

x^4 + x^3 - 5x^3 - 5x^2 + 9x^2 - 9 = 0

x^3(x + 1) - 5x^2(x + 1) + 9(x - 1)(x + 1) = 0

(x+1)(x^3 - 5x^2 + 9x - 9) = 0

(x+1)(x^3 - 3x^2 - 2x^2 + 6x + 3x - 9) = 0

(x+1)(x^2(x - 3) - 2x(x - 3) + 3(x - 3)) = 0

(x+1)(x-3)(x^2 - 2x + 3) = 0

x=-1

x=3

два действительныъ корня

x^2 - 2x + 3 = 0

D = 4 - 12 = -8

x12=(2+-2√2*i)/2 = 1 + - i√2 два комплексных