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

• This polynomial has surely one root, because its degree is odd. However, it has no rational roots. – Crostul Jan 25 '17 at 17:17
• Well, there aren't rational roots. The derivative is a bit easier to work with, and you can find the four critical points. That certainly helps. – lulu Jan 25 '17 at 17:18
• This function has 3 real root and inequlity mainly poroved or $x>\dfrac32$. – Nosrati Jan 25 '17 at 17:22
• – Arnaldo Jan 25 '17 at 17:24

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 $-1.8$.
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. 