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(B) Since $p(x)$ has a local maximum at $x=1$ and a local minimum at $x=3$, its derivative $p^{\prime}(x)$ must have roots at $x=1$ and $x=3$. Thus, $

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$9$

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(B) Since $p(x)$ has a local maximum at $x=1$ and a local minimum at $x=3$, its derivative $p^{\prime}(x)$ must have roots at $x=1$ and $x=3$. Thus, $p^{\prime}(x) = a(x-1)(x-3)$ for some constant $a 
eq 0$. Integrating $p^{\prime}(x)$, we get $p(x) = a(\frac{x^3}{3} - 2x^2 + 3x) + b$. Using $p(1) = 6$: $a(\frac{1}{3} - 2 + 3) + b = 6 \implies \frac{4a}{3} + b = 6 \implies 4a + 3b = 18$. Using $p(3) = 2$: $a(\frac{27}{3} - 2(9) + 3(3)) + b = 2 \implies a(9 - 18 + 9) + b = 2 \implies b = 2$. Substituting $b = 2$ into $4a + 3b = 18$, we get $4a + 6 = 18 \implies 4a = 12 \implies a = 3$. Thus, $p^{\prime}(x) = 3(x-1)(x-3)$. Evaluating at $x=0$, $p^{\prime}(0) = 3(0-1)(0-3) = 3(-1)(-3) = 9$.

Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$, then $p^{\prime}(0)$ is equal to

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