# Finding all solutions to an inequality equation

I have the following inequality that I need to find all solutions of:

$2x^3-8x > 5x^2-20$

My guess is that you would have to turn this into a polynomial equation and let the right hand side equal to $0$ (i.e. $2x^3-5x^2-8x+20=0$). By using the factor theorem you could guess a solution that is a factor of $20$, then use long division to solve for the other two roots. But how would you know whether the inequality is greater than the root (i.e. $>$) or less than the root (i.e. $<$)? Is it something you just need to guess and check? Or is there another way?

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I think you can use that $x=2$ is a root of your equation to help you factor the lefthand side, and then use a sign chart to determine where the left side is positive. – user84413 Aug 3 '13 at 22:46

Notice that both sides factor:

$$2x(x-2)(x+2)>5(x-2)(x+2)$$

$$(2x-5)(x-2)(x+2)>0$$

The roots are thus $-2,2,$ and $\frac{5}{2}$. Since this is a positive cubic, we know it approaches infinity as $x$ gets large, so we must have $x>\frac{5}{2}$ as possible solutions. Next, notice that none of the roots are double roots, so the polynomial will change sign at each. This means that it is negative in the range $(2,\frac{5}{2})$, positive in the range $(-2,2)$, and negative in the range $(-\infty,-2)$. So the answer is:

$$(-2,2)\cup(\frac{5}{2},\infty)$$

Notice I've done just about what you recommended, except factoring made finding the roots easier, and I didn't need to test any points because of the shape of a cubic polynomial with no double roots.

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Once you have found the roots of the polynomial, you can write the inequality in the form: $$(x-a)(x-b)(x-c)>0,$$ with $a\leq b \leq c$. The real line is so divided in 4 intervals:$$\mathbb R=(-\infty,a] \cup (a,b]\cup (b,c]\cup (c,+\infty).$$ Now all you need to do is to check the sign of the polynomial in these four regions.

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@Ryan Actually, you only need to check it in one region. The sign will alternate between regions. – Ataraxia Aug 3 '13 at 22:48
This is not true in the general case: take $$p(x)=\frac{1}{3}x^3+x^2-\frac{4}{3},$$ which is zero in $-2$ and negative in all neighbourhood of $-2$. It is possible to have a maximum in a root. – pppqqq Aug 3 '13 at 23:13
Well that's the case when a root has an even multiplicity, in which case I take there to be an implicit "interval" (-2,-2). Guess I should have clarified that point. – Ataraxia Aug 3 '13 at 23:34