Let the quadratic polynomial be \( f(x) = a x^2 + b x + c \).

Given that $$ f(-1) + f(2) = 0 $$ and one of the roots of \( f(x) = 0 \) is \(3\).

Let the other root be \(\alpha\). Hence, \( f(x) = k(x - 3)(x - \alpha) \), where \(k \ne 0\).

Substitute the condition \( f(-1) + f(2) = 0 \): $$ k(-1 - 3)(-1 - \alpha) + k(2 - 3)(2 - \alpha) = 0 $$ Simplify: $$ (-4)(-1 - \alpha) + (-1)(2 - \alpha) = 0 $$ $$ 4(1 + \alpha) - 2 + \alpha = 0 $$ $$ 2 + 5\alpha = 0 \Rightarrow \alpha = -\frac{2}{5}. $$
Therefore, the other root is \(-\dfrac{2}{5}\), which lies in $$\boxed{(-1,\ 0)}.$$