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Homesolving certain polynomial inequalities
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solving certain polynomial inequalities
In this article, we discuss inequality of the form $p(x)\geq 0$ or $p(x)>0$, where $p(x)$ is a polynomial with real coefficients such that $p(x)$ can be expressed as product of linear factors:
$p(x)=(xa_{1})(xa_{2})\cdots(xa_{n})$ 
where $a_{i}$ are real numbers (in the language of field theory, this means that $p(x)$ splits in the field of real numbers).
When we plot the polynomial $y=p(x)$, whenever it crosses the $x$axis, the crossing point is a root of $p(x)$. On either side of the crossing point, the values of $p(x)$ may be negative or positive. The way to solve the inequality $p(x)>0$ or $p(x)\geq 0$ is to look at how $p(x)$ crosses the $x$axis, and to realize that when $x$ is very large, $p(x)$ will be positive. This idea can be illustrated in the following figure:
If we start from the far right, the curve (graph of $p(x)$) is above the horizontal axis, and so the values of $p(x)$ are positive there. As it approaches $a_{1}$, its values decrease until it crosses $a_{1}$ and dips below the horizontal axis. The values of $p(x)$ are now negative. As it continues to travel along to the left, its values increase until it passes over another crossing point, $a_{2}$, and the values become positive as soon as it passes $a_{2}$. When $p(x)$ reaches $a_{3}$ however, it merely touches the $x$axis (at $a_{3}$) and then rises again. So on either side of $a_{3}$ the values of $p(x)$ are positive. Nevertheless, an analysis of how $p(x)$ crosses the $x$axis is enough to give us some clue on how to solve inequalities of the form $p(x)\geq 0$ or $p(x)>0$.
With this idea in mind, the steps are devised when solving inequalities of this type:
1. arrange $a_{i}$ so that they are in the ascending order: $a_{1}\geq a_{2}\geq\cdots\geq a_{n}$
2. 3. 4. go to point $a_{2}$
5. if $a_{2}\neq a_{1}$, label above the interval $i_{2}:=(a_{2},a_{1})$ adjacent to $i_{1}$ negative
6. if $a_{2}=a_{1}$, label above $a_{2}$ negative
7. 8. stop when $(\infty,a_{n})$ is labeled.
9. if we are trying to solve $p(x)\geq 0$, then all intervals that are labeled positive, including the end points, are solutions to the inequality
10. if we are solving $p(x)>0$, then all intervals labeled positive, excluding the end points, are solutions to the inequality.
Remark. After all the labeling is done, there should a total of $n+1$ labels, one over each interval, including the null intervals (the points).
To see how this works, let us look at some actual examples.

Solve $(x2)x(x+3)\geq 0$.
(a) Plot $2$, $0$, $3$ on the number line.
(b) The intervals separated by these points are $(2,\infty),(0,2),(3,0),(\infty,3)$.
(c) Since no two points are the same, the intervals that are labeled positive are $(2,\infty)$ and $(3,0)$.
(d) The solutions to the inequality are $[2,\infty)\cup[3,0]$, the square brackets signify that the end points are included in the solutions.

Solve $(x8)^{7}>0$.
(a) Plot $8$ on the number line.
(b) The intervals separated by these points are $(8,\infty),(\infty,8)$, since $8$ is repeated $7$ times.
(c) Start with labeling $(8,\infty)$ positive (the first label)
(d) The point $8$ is then labeled alternately $(2),+(3),(4),+(5),(6),+(7)$.
(e) The last label goes to $(\infty,8)$, which is negative.
(f) Therefore, the solution set is $(8,\infty)$ (excluding $8$).

Solve $(x1)(x+1)^{4}\geq 0$.
(a) Plot $1,1$ on the number line.
(b) The intervals separated by these points are $(1,\infty),(1,1)$, and $(\infty,1)$, since $1$ is repeated $4$ times.
(c) Start with labeling $(1,\infty)$ positive (the first label), followed by $(1,1)$ as negative.
(d) The point $1$ is then labeled alternately $+(3),(4),+(5)$.
(e) The last label goes to $(\infty,1)$, which is negative.
(f) Therefore, the solution set is $[1,\infty)\cup\{1\}$.
Remarks.
1. In the last two examples, we observe that a simplification can be made when solving the inequality: whenever we have repeating roots ($a_{i}=a_{{i+1}}$). Depending on the parity of the number of repeating roots, we have two situations:

If the number $n_{i}$ of repeating root, say $a_{i}$, is odd, then solving inequality involving $(xa_{i})^{{n_{i}}}$ is the same as solving the same inequality with $(xa_{i})^{{n_{i}}}$ replaced by $(xa_{i})$. In other words, their solution sets are the same. For example,
solving $(x8)^{7}>0$ is the same as solving $(x8)>0$.

If the number $n_{i}$ of repeating root $a_{i}$ is even, then we look at whether the inequality is strict or not.

If the inequality is strict, then we can completely eliminate $(xa_{i})^{{n_{i}}}$ from the inequality without altering the solution set. For example,
solving $(x1)(x+1)^{4}>0$ is the same as solving $(x1)>0$.

Otherwise, we need to remember the roots themselves as solutions. Therefore, the solution set for $(x1)(x+1)^{4}\geq 0$ is the same as the solution set of $(x1)\geq 0$ together with the root $4$.


2. Using the rules above, we may also solve inequalities $p(x)<0$ or $p(x)\leq 0$. The solution set for $p(x)<0$ is the complement of the solution set for $p(x)\geq 0$, and the solution set for $p(x)\leq 0$ is the complement of the solution set for $p(x)>0$.
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