<s>[INST] Confused by the solution of $x^3+bx^2+cx+d=0$ From $x^3 + bx^2 + cx + d = 0$, we have $(x-x_1)(x-x_2)(x-x_3)=0$ for some roots $x_1$, $x_2$ and $x_3$.
Expanding this second expression gives us
$$x^3 + \left(x_1+x_2+x_3\right)x^2 + \left(x_1x_2 + x_1 x_3 + x_2 x_3\right) x + x_1 x_2 x_3 = 0$$
and comparing coefficients gives


*

*$x_1+x_2+x_3 = b$

*$x_1 x_2 + x_1 x_3 + x_2 x_3 = c$

*$x_1 x_2 x_3 = d$


From this we can see that


*

*$x_2 x_3 = d/x_1$ (where $x_1 \neq 0$)

*$x_2 + x_3 = b - x_1$

*$x_1 \left(x_2 + x_3\right) + x_2 x_3 = c$


This gives us
$$x_1 \left(b-x_1\right) + \frac{d}{x_1} = c \implies -x_1^3 + bx_1^2 - cx_1 + d = 0
\implies x_1^3 - bx_1^2 + cx_1 -d = 0$$
But since $x_1$ is a root of the polynomial, we know that
$$x_1^3 = bx_1^2 + cx_1 + d = 0$$
and differencing the two gives
$$bx_1^2 + d = 0$$
so we have $x_1 = \pm \sqrt{-d/b}$, but that’s not always true. Why?
 [/INST] If $$x^3+bx^2+cx+d=(x-x_1)(x-x_2)(x-x_3),$$
then 
$$x_1+x_2+x_3=-b$$
$$x_1x_2x_3=-d$$
instead of
$$x_1+x_2+x_3=b$$
$$x_1x_2x_3=d$$
because
$$(x-x_1)(x-x_2)(x-x_3)$$
$$=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_1)x-x_1x_2x_3.$$
</s>