How to solve ${x^5-3x^3+2x^2-2>0}$ without using a graphing calculator? I was just wondering if I'm able to solve quickly the following inequality without using a graphing calculator:
${x^5-3x^3+2x^2-2>0}$
Any tips?
EDIT 1: scientific calculator is allowed
 A: The roots are not rational, so you are in for numeric solution.The three real roots are about $-2, -0.75,$ and $+1.5$  The $x^5$ term will dominate when you get more than $\pm 3$, so I would start by computing the values at $-2,-1,0,1,2$  This turns out to bracket the roots and you can use your favorite root finder.  For calculator use bisection is not so bad, probably tweaked to use shorter decimals and with some bias based on the previous results.  For example we have $f(-2)=-2, f(-1)=+2, f(-1.5)\approx 5$ so my next try for the root near $-2$ might be $-1.8$.
A: I suspect the main thing they want you to think about is
$$ x^5 - 3 x^3 + 2 x^2 - 2 = x^2 (x-1)^2 (x+2) - 2. $$
This tells us that $x=0,1$ are critical points, actually local minima. There is a local maximum between them, as well as two of the inflection points.
The thing to do with a scientific calculator is to graph the thing by hand, using numbers provided by the calculator. Also check for reasonability; I made one mistake, had to start over.
This is an HP-15C, programmable. I put three named programs, A was the function, B was the first derivative, C was the second derivative. The calculator also searches for a root of a function, given two endpoints in which a root lies. I think it uses the secant method. 
For people who have no graph paper, https://www.printablepaper.net/category/graph

