Question

Calculati in Z
(x+3)^2+[(x+3)(x-2)]^2=0
x=?

Answer 1

$x {}^{2} + 6x + 9 + (x {}^{2} - 2x + 3x - 6) {}^{2} = 0$

$x {}^{2} + 6x + 9 + (x {}^{2} + x - 6) {}^{2} = 0$

$x {}^{2} + 6x + 9 + x {}^{4} + x {}^{2} + 36 + 2x {}^{3} - 12x {}^{2} - 12x = 0$

$x {}^{4} + {2x}^{3} - 10x {}^{2} - 6x + 45 = 0$

$x {}^{3} (x + 3) - x {}^{2} (x + 3) - 7x(x + 3) + 15(x + 3) = 0$

$(x + 3)(x {}^{3} - x {}^{2} - 7x + 15) = 0$

$(x + 3)(x + 3)(x {}^{2} - 4x + 5) = 0$

$(x + 3) {}^{2} = 0$

$x = - 3$