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(C) Let the quadratic polynomial be $f(x) = a(x - 3)(x - \alpha)$,where $\alpha$ is the other root.Given $f(2) = a(2 - 3)(2 - \alpha) = a(-1)(2 - \alp

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$(-1, 0)$

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(C) Let the quadratic polynomial be $f(x) = a(x - 3)(x - \alpha)$,where $\alpha$ is the other root.Given $f(2) = a(2 - 3)(2 - \alpha) = a(-1)(2 - \alpha) = a(\alpha - 2)$.Given $f(-1) = a(-1 - 3)(-1 - \alpha) = a(-4)(-1 - \alpha) = 4a(1 + \alpha)$.Since $f(-1) + f(2) = 0$,we have $4a(1 + \alpha) + a(\alpha - 2) = 0$.Since $a 
eq 0$,we can divide by $a$: $4 + 4\alpha + \alpha - 2 = 0$.$5\alpha + 2 = 0$ $\Rightarrow 5\alpha = -2$ $\Rightarrow \alpha = -\frac{2}{5} = -0.4$.Thus,the other root $\alpha = -0.4$ lies in the interval $(-1, 0)$.

Let $f(x)$ be a quadratic polynomial such that $f(-1)+f(2)=0$. If one of the roots of $f(x)=0$ is $3$,then its other root lies in

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