You are here
Homesolving certain polynomial inequalities
Primary tabs
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 $p(x)>0$.
Mathematics Subject Classification
12D99 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections