For the product to be negative, one
but not both of the factors must be negative.
Where is 3x^2 - 10x + 3 less than zero? Assuming 3x^2 - 10x + 3 < 0, where is 2x^2 - 5x + 2 greater than zero? What is the overlap of these solutions?
. . . . .$\displaystyle 3x^2\, -\, 10x\, +\, 3\, <\, 0$
. . . . .$\displaystyle (3x\, -\, 1)(x\, -\, 3)\, <\, 0$
From the equality, (3x - 1)(x - 3) = 0, we know that the above crosses the x-axis at x = 1/3 and at x = 3. From the graph of y = 3x^2 - 10x + 3, we know that the graph is below the x-axis (and thus that the inequality is negative) between these zeroes; that is, on the interval (1/3, 3).
. . . . .$\displaystyle 2x^2\, -\, 5x\, +\, 2\, >\, 0$
. . . . .$\displaystyle (2x\, -\, 1)(x\, -\, 2)\, >\, 0$
From the equation, (2x - 1)(x - 2) = 0, we know the graph crosses the x-axis at x = 1/2 and at x = 2. From the graph of y = 2x^2 - 5x + 2, we know that the graph is above the x-axis (and thus that the inequality is positive) "on the ends"; that is, on the intervals (-infinity, 1/2) and (2, +infinity).
For both inequalities to have the signs we want, we need the intersection of these two solutions: [(-infinity, 1/2) union (2, +infinity)] intersect (1/3, 3).
What then is the solution-interval for this case?
Now what if you start with the inequalities on the factors going in the other direction?