Solve graphically (x-2)^3(x+4)(3-x)^2(x+1)^4 <0

Answer 1

You know where the roots are.
You know that if the root is of odd order, the graph crosses the x-axis there.
And, if the root is of even order, the graph is just tangent there.

So, knowing what you do about the general shape of polynomials, since this one is of even order,

The roots are -4,-1,2,3

y > 0 for -∞ < x < -4
y < 0 for -4 < x < -1
y < 0 for -1 < x < 2
y > 0 for 2 < x < 3
y > 0 for 3 < x < ∞

Verify this at

http://www.wolframalpha.com/input/?i=(x-2)%5E3(x%2B4)(3-x)%5E2(x%2B1)%5E4+,+-2+%3C+x+%3C+10%2F3