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(A) Let $f(x) = ax^3 + bx^2 + cx$. Since $f(x)$ is a polynomial,it is continuous and differentiable on $\mathbb{R}$. We observe that $f(0) = a(0)^3 +

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At least one root in $[0, 1]$

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(A) Let $f(x) = ax^3 + bx^2 + cx$. Since $f(x)$ is a polynomial,it is continuous and differentiable on $\mathbb{R}$. We observe that $f(0) = a(0)^3 + b(0)^2 + c(0) = 0$. Given $a + b + c = 0$,we have $f(1) = a(1)^3 + b(1)^2 + c(1) = a + b + c = 0$. Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $\alpha \in (0, 1)$ such that $f'(\alpha) = 0$. Calculating the derivative,$f'(x) = 3ax^2 + 2bx + c$. Thus,$f'(\alpha) = 3a\alpha^2 + 2b\alpha + c = 0$ for some $\alpha \in (0, 1)$. Therefore,the quadratic equation $3ax^2 + 2bx + c = 0$ has at least one root in $[0, 1]$.

If $a, b, c$ are real numbers such that $a + b + c = 0$,then the quadratic equation $3ax^2 + 2bx + c = 0$ has

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