$$x^3+bx^2+cx+d =0$$

In the equation above, $b, c,$ and $d$ are constants. If the equation has roots $-2,0, \text{ and } 4$, what is the value of $b+c+d$ ?

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Given the roots, we can write the equation of the polynomial in factored form:

$$a(x+2)(x-0)(x-4)=0$$

Expanding:

$$ax(x+2)(x-4)=0$$ $$ax(x^2-2x-8)=0$$ $$a(x^3-2x^2-8x)=0$$

$a$ must equal zero for this equation to match the given equation. Comparing our equation with the given:

$$x^3+bx^2+cx+d = x^3-2x^2-8x$$ $$b=-2, c=-8, d=0$$ $$b+c+d = -2+(-8)+0$$ $$= \boxed{-10}$$