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(C) Given that $p^{\prime}(x) > 0$ for all $x \in \mathbb{R}$, the polynomial $p(x)$ is a strictly increasing function on the entire real line. Since

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$p(x)$ may have a negative real root

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(C) Given that $p^{\prime}(x) > 0$ for all $x \in \mathbb{R}$, the polynomial $p(x)$ is a strictly increasing function on the entire real line. Since $p(x)$ is strictly increasing, it can have at most one real root. We are given $p(0) = 1$. Since $p(x)$ is strictly increasing and $p(0) = 1 > 0$, for any $x > 0$, $p(x) > p(0) = 1$, so $p(x)$ cannot have any positive real roots. As $x 	o -\infty$, $p(x) 	o -\infty$ (assuming the degree is odd) or $p(x) 	o c$ (if degree is even, but $p^{\prime}(x) > 0$ implies the degree must be odd for the limit to be $-\infty$). Since $p(0) = 1$ and the function is continuous and strictly increasing, by the Intermediate Value Theorem, there must exist some $x_0 Thus, $p(x)$ has exactly one negative real root.

Let $p(x)$ be a polynomial with real coefficients, $p(0) = 1$ and $p^{\prime}(x) > 0$ for all $x \in \mathbb{R}$. Then

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