Why do you have to check all the 'sections' of an inequality when solving for x? For instance, the inequality
$x^2+3x+2>0$ factors into $(x+2)(x+1)>0$
If this were just an equation (i.e ...=0) you would know the solutions are x={-2, -1}.
But, because it's an inequality, you have to check the intervals. For instance, by drawing a number line and evaluating the inequality expression at each of the intervals and marking whether it satisfies the condition or not. How come the usual method doesn't work?
If you use the inequality symbol is an equals sign and do it the usual way you would get $x>-1$ and $x>-2$ which is not right.
 A: So, we know that
$$(x+2)(x+1) > 0$$
from the outset, just by factoring it. Nothing special is going on there.
But now consider: there are two cases when a product $ab$ is greater than $0$ ($ab>0$):

*

*$a > 0$ and $b > 0$

*$a<0$ and $b<0$
Applying the first case to our quadratic gives
$$x+2 > 0 \text{ and } x+1 > 0$$
Which $x$ values satisfy both inequalities? Clearly, it will be all $x > -1$.
Applying the second one gives
$$x+2 < 0 \text{ and } x+1 < 0$$
Again, which $x$ satisfy both inequalities? Clearly, it will be $x<-2$
Hence, the set of desired $x$ is all $x$ such that $x<-2$ or $x > -1$. Rewritten, this is
$$(-\infty,-2) \cup (-1,+\infty)$$

Your method fails in two respects.
Firstly, it assumes that both factors must be positive. A negative sign is also possible.
Secondly, your method has a solution set of $x>-1$ or $x>-2$. Notice how this is a bit strange? Certainly, $x=-1.5$ is covered by your method, but it is not actually a solution. (Just plug it back in.) This pokes at a critical linguistic item in mathematics: the difference between "or" and "and".
If $x$ is to satisfy $x+2 > 0$ and $x+1>0$, and thus work in the first case, we need both (not just one!) to be satisfied. Thus, we have to consider what $x$ elements are common to both solution sets. Luckily, all $x > -1$ satisfy $x > -2$, hence why we use that "smaller" set of numbers.
A: In your example, the term that's compared against $0$ is fully factored. Now we have that $$ab=0\quad\iff\quad (a=0\lor b=0)\tag 1$$
whereas in the case of in inequality, we have
$$ab>0\quad\iff\quad ((a>0\land b>0)\lor (a<0\land b<0))\tag2 $$
so it's obvious that in case $(2)$ there is more work to be done than for case $(1)$.
More specifically, if case $(1)$ was made of $n$ factors, then you'll have to $\lor$ just $n$ factors, whereas in case $(2)$ there will be $2^{n-1}$ terms to be $\lor$'ed, each one consisting of $n$ terms to be $\land$'ed$^1$.
Notice that you can still build on $(1)$ to find solutions of $(2)$ as follows, provided the left-hand side is continuous:

*

*Find all solutions to $f(x) = 0$.  The solutions be $n$ intervals. In most cases, the intervals are degenerate and consist of just one single point.  In the example, there are two such "intervals", namely $[-1]$ and $[-2]$.


*Remove all intervals found in step 1 from the real line $\Bbb R$. This will result in $n+1$ proper intervals.  In your example, there will be 3 intervals $I_1=(-\infty,-2)$, $I_2=(-2,-1)$ and $I_3=(-1,\infty)$.


*For each interval $I_k$ from step 2 pick some $x_k\in I_k$. Then

*

*$f(x_k) > 0$ iff $f(x)>0$ for all $x\in I_k$.

*$f(x_k) < 0$ iff $f(x)<0$ for all $x\in I_k$.



The case when in step 1 a complete interval is found with $f(x) = 0$ can also occur.  A trivial example is $f(x) = 0$.  A non-trivial example is $f(x) = |x| - x $.
$^1$Depending on the anatomy of the product you might eleminate most of these cases, though.
A: You can alternatively tackle the problem as follows:
\begin{align*}
(x + 2)(x + 1) > 0 & \Longleftrightarrow ((x + 1) + 1)(x + 1) > 0\\\\
& \Longleftrightarrow (x + 1)^{2} + (x + 1) > 0 \\\\
& \Longleftrightarrow (x + 1)^{2} + (x + 1) + \frac{1}{4} > \frac{1}{4}\\\\
& \Longleftrightarrow \left[(x + 1) + \frac{1}{2}\right]^{2} > \frac{1}{4}\\\\
& \Longleftrightarrow \left(x + \frac{3}{2}\right)^{2} > \frac{1}{4}\\\\
& \Longleftrightarrow \left|x + \frac{3}{2}\right| > \frac{1}{2}\\\\
& \Longleftrightarrow \left(x < -2\right)\vee(x > -1)
\end{align*}
Hopefully this helps!
