^That's exactly it!
Factoring isn't going to happen
Actually, I'm retarded. This thing factors nicely into (3*x + 4)*(x^2 + 2*x - 5). From there all you need to do is figure out when x^2+2x-5 equals zero, because that will drive the entire expression to zero regardless of what the first part is (we already saw that one of the solutions is -4/3, which drives that part to zero). Anyhow, using the quadratic equation gives you an irrational pair of solutions, -1 +/- sqrt(6).
Edit: Well, I'm sick and heading to bed early. To summarize the second problem:
Given problem: 3*x^3 + 10*x^2 - 7*x - 20 = 0
Factor into: (3*x + 4)*(x^2 + 2*x - 5) = 0
Set each part equal to zero:
(3*x + 4) = 0
(x^2 + 2*x - 5) = 0
AND solve for x in each case. The first one is straightforward and gives you x = -4/3. The second one is probably best handled by smashing it with the quadratic formula, which you ought to know well when taking any algebra course
This gives you x = -1 +/- sqrt(6), where "sqrt" means square root.
x = -4/3, -1 + sqrt(6), -1 - sqrt(6)