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(B) Given $f(x)$ is negative in $\left(-\infty, -\frac{5}{3}\right) \cup (3, \infty)$ and positive in $\left(-\frac{5}{3}, 3\right)$,we can write $f(x

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positive in $[0, 3) \cup \left(3, \frac{9}{2}\right)$ and negative in $\left(\frac{9}{2}, 5\right]$

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(B) Given $f(x)$ is negative in $\left(-\infty, -\frac{5}{3}\right) \cup (3, \infty)$ and positive in $\left(-\frac{5}{3}, 3\right)$,we can write $f(x) = -k_1(x + \frac{5}{3})(x - 3)$ where $k_1 > 0$. This simplifies to $f(x) = k_1(x + \frac{5}{3})(3 - x)$.Given $g(x)$ is negative in $\left(3, \frac{9}{2}\right)$ and positive elsewhere,we can write $g(x) = k_2(x - 3)(x - \frac{9}{2})$ where $k_2 > 0$.Now,consider the product $P(x) = f(x)g(x) = k_1 k_2 (x + \frac{5}{3})(3 - x)(x - 3)(x - \frac{9}{2})$.Since $(3 - x) = -(x - 3)$,we have $P(x) = -k_1 k_2 (x + \frac{5}{3})(x - 3)^2 (x - \frac{9}{2})$.Let $K = k_1 k_2 > 0$. Then $P(x) = -K (x + \frac{5}{3})(x - 3)^2 (x - \frac{9}{2})$.The roots are $x = -\frac{5}{3}, 3, \frac{9}{2}$. Note that $x = 3$ is a root of multiplicity $2$,so the sign does not change across $x = 3$.For $x > \frac{9}{2}$,$P(x)$ is negative. For $3  -\frac{5}{3}$),$P(x)$ is positive.Thus,in the interval $[0, 5]$,$P(x)$ is positive in $[0, 3) \cup (3, \frac{9}{2})$ and negative in $(\frac{9}{2}, 5]$. The correct option is $B$.

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