How to solve a polynomial inequality? Ok guys, I need some more help with a question for my girlfriend. Basically she was given a problem on a test/quiz and the only way I know how to do it is with a method that she hasnt learned in class yet. So pretty much I want you guys to look it over and let me know if there is another method to the problem that she might know. The problem is...
$$2x^3 + 3x^2 \lt 11x + 6$$
She has to solve it and give the answer in interval notation. So I would first move everything from the right side to the left to have a third order polynomial. My next step would be to use the rational zero test to start finding a zero using synthetic division. After I found one zero, I would factor it out of the function, and i would be left with $(x-a)$*2nd orderpolynomial. I could factor out the polynomial and find my 3 zero's. 
The only problem is, she didnt learn synthetic divison, rational zero test, or long division of polynomials. is there another way she could do this problem with knowledge she might have? i tried to briefly teach her my way and it went over her head
 A: Factor like this
$$2x^3 +3x^2 - 11x - 6 = (x-2)(2x+1)(x+3).$$
Then draw a sign chart.
A: As has been noted, the easiest way to solve a polynomial inequality $f(x) < 0$ (or $f(x) > 0$) is to find the roots of the polynomial $f$, and then form a sign chart.
I would like to mention the reason why sign charts work.  In my limited experience as a teacher, knowing the reason why a certain methods works is usually just as important as knowing the method itself.
First, note that polynomials are very well behaved.  In particular, polynomials are continuous.  Intuitively, this means that in order to sketch the function $f(x)$, one does not need to lift their pencil from the paper.  In other words, the function has no sudden jumps or holes.  Consequently, the places where the function crosses the $x$-axis (i.e. the places where $f(x) = 0$ are exactly the places where the function can change sign.  Therefore, solving polynomial inequalities amounts to finding the zeroes of the function and determining the sign of each interval.  However, finding the sign of each interval is also straight-forward since every point in any given interval will have the same sign.  Therefore, one just has to choose a test-point for each interval to determine the sign of the entire interval (which in turn solves inequality).
A: You could just move everything to one side and graph it to find approximate roots, then substitute them in.  Or you could feed it to Wolfram Alpha
