Expand the product: $(x-3)(x+2)=x^2+2x-3x-6=x^2-x-6$. In standard form, the constant term is the $y$-intercept, so this form shows directly that the graph crosses the $y$-axis at $-6$.

Since $p(x)=3x+r$, substitute $x=2$ into the expression for the polynomial:

Because $p(2)=11$, we have $6+r=11$, so $r=5$.

A polynomial has factor $x-r$ when $p(r)=0$. From the table, $p(2)=0$, so $x-2$ must be a factor of $p(x)$.

If $x-4$ is a factor, then the polynomial equals $0$ when $x=4$. So $x=4$ must be a solution to $x^2-7x+12=0$.

If a point on the $x$-axis has coordinate $x$, then its distances to the intercepts at $2$ and $7$ are represented by $|x-2|$ and $|x-7|$. An expression that could represent the rectangle's area is the product of those distances. Among the choices, $(x-2)(x-7)$ matches the standard polynomial built from the two intercepts and could represent that area for values of $x$ outside the interval from $2$ to $7$.