Confused by the solution of $x^3+bx^2+cx+d=0$

Question

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

1. $x_1+x_2+x_3 = b$
2. $x_1 x_2 + x_1 x_3 + x_2 x_3 = c$
3. $x_1 x_2 x_3 = d$

From this we can see that

1. $x_2 x_3 = d/x_1$ (where $x_1 \neq 0$)
2. $x_2 + x_3 = b - x_1$
3. $x_1 \left(x_2 + x_3\right) + x_2 x_3 = c$

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?

Answer 1

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 $$(x-x_1)(x-x_2)(x-x_3)$$ $$=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_1)x-x_1x_2x_3.$$