Answer
The possible real solutions of the equation of
$x^3+4x^2+4x+3=0$ is:
$x=-3$
Work Step by Step
$x^3+4x^2+4x+3=0$
$x^3+x^2+x+3x^2+3x+3=x(x^2+x+1)+3(x^2+x+1)=(x+3)(x^2+x+1)=0$
$(x+3)(x^2+x+1)=0$
The second term can not be further evaluated, it hasn't got any real roots.
The possible real solutions of the equation of
$(x+3)(x^2+x+1)=0$ is:
$x=-3$
