Solving inequality $x^3-4x>0$ I was sort of finding the roots by doing $x^3-4x>0=x(x^2-4)$
$x = 0, x = -2,x = 2$ for $$x(x-2)(x+2)>0$$ 
Then I stopped and thought; maybe I shouldn't be doing that? I am doubting myself! Can anyone confirm if I am doing the right thing to solve the equality? 
 A: I like to do tables to determine signs. It follows from Bolzano's Theorem the sign will be kept between the roots. You already factored this and obtained $x=2,x=-2,x=0$.
$$\begin{array}{|c|c|c|c|} 
 & (-\infty,-2) & (-2,0)& (0,2)&(2,+\infty)\\ \hline
x-2 & - &- & -&+\\  \hline
x+2 &- &+  &+ &+\\ \hline
x & - & - &+ &+\\ \hline
p(x)& -&+&-&+\\ \hline
\end{array}$$
Explanation: The top row is the real line divided into the intervals by the roots we have. The first three rows of $+/-$ account for the sign of the factors in each interval, which we obtain by inspection. Finally, the sign of $p$ is obtained by "multiplying" each of the values obtained, so $-\times-\times-=-$, $-\times+\times +=+$,$\&c$.
Try and do the same for $p(x)={x^3} + 4{x^2} + 3x - 2$.
A: A good way to do this is to draw a number line and mark the places where the polynomial is zero. The sign will be constant in the intervals so created.  Check inside each one to see the sign (+ or -).  Then it's easy to read the solution.
A: The polynomial $x^3-4x$ is $0$ at $x=-2, 0,$ and $2.$ At $-3$ the value of the polynomial is $< 0$; at $3$ the value is $>0.$ At $-1,$ the value is $>0.$ At $1$ the value is $<0.$ 
From this we see the polynomial is $> 0$ for
$$-2<x<0 \text{ and } x>2.$$
A: This is a good approach.  You are almost there.  A product is positive if it has an even number of negative terms.  So it is positive if there are no negative terms, which is if $x \gt 2$, or if there are two negative terms ...
