Answer: intercepts \((-1,0)\), \((2,0)\), \((3,0)\), \((0,-6)\); \(x\lt -1\) or \(2\lt x\lt 3\).

The \(x\)-intercepts are found from

\((x-2)(x+1)(3-x)=0.\)

So

\(x=2,\qquad x=-1,\qquad x=3.\)

The \(y\)-intercept is found by putting \(x=0\):

\(y=(-2)(1)(3)=-6.\)

The cubic has leading term \(-x^3\), so the left-hand end is above the \(x\)-axis and the right-hand end is below the \(x\)-axis.

Using the intercepts in order, the expression is positive on

\(x\lt -1\)

and also on

\(2\lt x\lt 3.\)